题意

计算满足下列条件长度为 $n$ 的序列 $(a_1,a_2,\cdots ,a_n)$ 的个数。

• 对于 $1\le i\le n$ 满足 $0\le a_i\le 2^m$
• $\sum _{i=1}^n{\text{cnt}}_1(a_i\oplus b_i)=k$

其中序列 $b$ 为 $(a_2,a_3,\cdots ,a_n,a_1)$，${\text{cnt}}_1(x)$ 代表 $x$ 二进制中 $1$ 的个数。

对 $998244353$ 取模。

$2\le n<998244353,1\le m\le 10^8,1\le k\le 5\times 10^4$

分析：

首先观察第二个条件，由于每一位是独立的，考虑拆位。

统计 $m$ 位中每一位对答案的贡献，记 $a_i=\sum _{j=0}^{m-1}{c}_{ij}\times 2^j$，其中 $c_{ij}$ 表示 $a_i$ 二进制的第 $j$ 位。

那么对第 $j$ 位来说，只需要知道序列 $(c_{1j}\oplus c_{2j},c_{2j}\oplus c_{3j},\cdots ,c_{(n-1)j}\oplus c_{nj},c_{nj}\oplus c_{1j})$ 中产生若干个 $1$ 的方案数。

下面简单证一下序列中只可能包含偶数个 $1$：

$c_{pj}\oplus c_{qj}=1$ 等价于 $c_{pj}\ne c_{qj}$

$c_{pj}\oplus c_{qj}=0$ 等价于 $c_{pj}=c_{qj}$

所以问题等价于在 $01$ 环选择 $u$ 条边使相邻点的值不相等，那么剩下 $n-u$ 条边使得相邻点的值相等。

由于相等边不改变值，所以我们可以将不相等边构成的连通块缩成一个连通块，也就是将相等边构成连通块的两边节点合并到一起。

此时问题变为二分图染色问题，我们知道二分图染色奇数环是不可行的，所以环大小必定为偶数，由此推出 $1$ 必须为偶数个。

那么得到 $2u$ 个 $1$ 的方案数就为 $\left(\genfrac{}{}{0ex}{}{n}{2u}\right)$，考虑每一位的生成函数

$$F(x)=\sum _{u=0}^{\lfloor \frac{n}{2}\rfloor }\left(\genfrac{}{}{0ex}{}{n}{2u}\right)x^u$$

由于每一位独立，所以 $m$ 个生成函数都相等，所以答案为

$$[x^k]F^m(x)$$

由于 $n<998244353$ ，但 $k\le 5\times 10^4$，所以 $F(x)$ 至多算到第 $k$ 项，那么 $\left(\genfrac{}{}{0ex}{}{n}{2u}\right)$ 经典维护下降幂。最后使用多项式快速幂求解即可。

代码：

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
constexpr int mod = 998244353;
template<class T>
T power(T a, int b) {T res = 1;for (; b; b /= 2, a *= a) {if (b % 2) {res *= a;}}return res;
}
template<int mod>
struct ModInt {int x;ModInt() : x(0) {}ModInt(i64 y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}ModInt &operator+=(const ModInt &p) {if ((x += p.x) >= mod) x -= mod;return *this;}ModInt &operator-=(const ModInt &p) {if ((x += mod - p.x) >= mod) x -= mod;return *this;}ModInt &operator*=(const ModInt &p) {x = (int)(1LL * x * p.x % mod);return *this;}ModInt &operator/=(const ModInt &p) {*this *= p.inv();return *this;}ModInt operator-() const {return ModInt(-x);}ModInt operator+(const ModInt &p) const {return ModInt(*this) += p;}ModInt operator-(const ModInt &p) const {return ModInt(*this) -= p;}ModInt operator*(const ModInt &p) const {return ModInt(*this) *= p;}ModInt operator/(const ModInt &p) const {return ModInt(*this) /= p;}bool operator==(const ModInt &p) const {return x == p.x;}bool operator!=(const ModInt &p) const {return x != p.x;}ModInt inv() const {int a = x, b = mod, u = 1, v = 0, t;while (b > 0) {t = a / b;swap(a -= t * b, b);swap(u -= t * v, v);}return ModInt(u);}ModInt pow(i64 n) const {ModInt res(1), mul(x);while (n > 0) {if (n & 1) res *= mul;mul *= mul;n >>= 1;}return res;}friend ostream &operator<<(ostream &os, const ModInt &p) {return os << p.x;}friend istream &operator>>(istream &is, ModInt &a) {i64 t;is >> t;a = ModInt<mod>(t);return (is);}int val() const {return x;}static constexpr int val_mod() {return mod;}
};
using Z = ModInt<mod>;
vector<Z> fact, infact;
void init(int n) {fact.resize(n + 1), infact.resize(n + 1);fact[0] = infact[0] = 1;for (int i = 1; i <= n; i ++) {fact[i] = fact[i - 1] * i;}infact[n] = fact[n].inv();for (int i = n; i; i --) {infact[i - 1] = infact[i] * i;}
}
Z C(int n, int m) {if (n < 0 || m < 0 || n < m) return Z(0);return fact[n] * infact[n - m] * infact[m];
}
vector<int> rev;
vector<Z> roots{0, 1};
void dft(vector<Z> &a) {int n = a.size();if (int(rev.size()) != n) {int k = __builtin_ctz(n) - 1;rev.resize(n);for (int i = 0; i < n; i ++) {rev[i] = rev[i >> 1] >> 1 | (i & 1) << k;}}for (int i = 0; i < n; i ++) {if (rev[i] < i) {swap(a[i], a[rev[i]]);}}if (int(roots.size()) < n) {int k = __builtin_ctz(roots.size());roots.resize(n);while ((1 << k) < n) {Z e = power(Z(3), (mod - 1) >> (k + 1));for (int i = 1 << (k - 1); i < (1 << k); i ++) {roots[i << 1] = roots[i];roots[i << 1 | 1] = roots[i] * e;}k ++;}}for (int k = 1; k < n; k *= 2) {for (int i = 0; i < n; i += 2 * k) {for (int j = 0; j < k; j ++) {Z u = a[i + j], v = a[i + j + k] * roots[k + j];a[i + j] = u + v, a[i + j + k] = u - v;}}}
}
void idft(vector<Z> &a) {int n = a.size();reverse(a.begin() + 1, a.end());dft(a);Z inv = (1 - mod) / n;for (int i = 0; i < n; i ++) {a[i] *= inv;}
}
struct Poly {vector<Z> a;Poly() {}Poly(const vector<Z> &a) : a(a) {}Poly(const initializer_list<Z> &a) : a(a) {}int size() const {return a.size();}void resize(int n) {a.resize(n);}Z operator[](int idx) const {if (idx < size()) {return a[idx];} else {return 0;}}Z &operator[](int idx) {return a[idx];}Poly mulxk(int k) const {auto b = a;b.insert(b.begin(), k, 0);return Poly(b);}Poly modxk(int k) const {k = min(k, size());return Poly(vector<Z>(a.begin(), a.begin() + k));}Poly divxk(int k) const {if (size() <= k) {return Poly();}return Poly(vector<Z>(a.begin() + k, a.end()));}friend Poly operator+(const Poly &a, const Poly &b) {vector<Z> res(max(a.size(), b.size()));for (int i = 0; i < int(res.size()); i ++) {res[i] = a[i] + b[i];}return Poly(res);}friend Poly operator-(const Poly &a, const Poly &b) {vector<Z> res(max(a.size(), b.size()));for (int i = 0; i < int(res.size()); i ++) {res[i] = a[i] - b[i];}return Poly(res);}friend Poly operator*(Poly a, Poly b) {if (a.size() == 0 || b.size() == 0) {return Poly();}int sz = 1, tot = a.size() + b.size() - 1;while (sz < tot) {sz *= 2;}a.a.resize(sz);b.a.resize(sz);dft(a.a);dft(b.a);for (int i = 0; i < sz; i ++) {a.a[i] = a[i] * b[i];}idft(a.a);a.resize(tot);return a;}friend Poly operator*(Z a, Poly b) {for (int i = 0; i < int(b.size()); i ++) {b[i] *= a;}return b;}friend Poly operator*(Poly a, Z b) {for (int i = 0; i < int(a.size()); i ++) {a[i] *= b;}return a;}Poly &operator+=(Poly b) {return (*this) = (*this) + b;}Poly &operator-=(Poly b) {return (*this) = (*this) - b;}Poly &operator*=(Poly b) {return (*this) = (*this) * b;}Poly deriv() const {if (a.empty()) {return Poly();}vector<Z> res(size() - 1);for (int i = 0; i < size() - 1; i ++) {res[i] = a[i + 1] * (i + 1);}return Poly(res);}Poly integr() const {vector<Z> res(size() + 1);for (int i = 0; i < size(); i ++) {res[i + 1] = a[i] / (i + 1);}return Poly(res);}Poly inv(int m) const {Poly x{a[0].inv()};int k = 1;while (k < m) {k *= 2;x = (x * (Poly{2} - modxk(k) * x)).modxk(k);}return x.modxk(m);}Poly log(int m) const {return (deriv() * inv(m)).integr().modxk(m);}Poly exp(int m) const {Poly x{1};int k = 1;while (k < m) {k *= 2;x = (x * (Poly{1} - x.log(k) + modxk(k))).modxk(k);}return x.modxk(m);}Poly pow(int k, int m) const {int i = 0;while (i < size() && a[i].val() == 0) {i ++;}if (i == size() || 1LL * i * k >= m) {return Poly(vector<Z>(m));}Z v = a[i];auto f = divxk(i) * v.inv();return (f.log(m - i * k) * k).exp(m - i * k).mulxk(i * k) * power(v, k);}Poly sqrt(int m) const {Poly x{1};int k = 1;while (k < m) {k *= 2;x = (x + (modxk(k) * x.inv(k)).modxk(k)) * ((mod + 1) / 2);}return x.modxk(m);}Poly mulT(Poly b) const {if (b.size() == 0) {return Poly();}int n = b.size();reverse(b.a.begin(), b.a.end());return ((*this) * b).divxk(n - 1);}vector<Z> eval(vector<Z> x) const {if (size() == 0) {return vector<Z>(x.size(), 0);}const int n = max(int(x.size()), size());vector<Poly> q(n << 2);vector<Z> ans(x.size());x.resize(n);function<void(int, int, int)> build = [&](int p, int l, int r) {if (r - l == 1) {q[p] = Poly{1, -x[l]};} else {int m = l + r >> 1;build(p << 1, l, m);build(p << 1 | 1, m, r);q[p] = q[p << 1] * q[p << 1 | 1];}};build(1, 0, n);function<void(int, int, int, const Poly &)> work = [&](int p, int l, int r, const Poly &num) {if (r - l == 1) {if (l < int(ans.size())) {ans[l] = num[0];}} else {int m = (l + r) / 2;work(p << 1, l, m, num.mulT(q[p << 1 | 1]).modxk(m - l));work(p << 1 | 1, m, r, num.mulT(q[p << 1]).modxk(r - m));}};work(1, 0, n, mulT(q[1].inv(n)));return ans;}Poly inter(const Poly &y) const {vector<Poly> Q(a.size() << 2), P(a.size() << 2);function<void(int, int, int)> dfs1 = [&](int p, int l, int r) {int m = l + r >> 1;if (l == r) {Q[p].a.push_back(-a[m]);Q[p].a.push_back(Z(1));return;}dfs1(p << 1, l, m), dfs1(p << 1 | 1, m + 1, r);Q[p] = Q[p << 1] * Q[p << 1 | 1];};dfs1(1, 0, a.size() - 1);Poly f;f.a.resize((int)(Q[1].size()) - 1);for (int i = 0; i + 1 < Q[1].size(); i ++) {f[i] = Q[1][i + 1] * (i + 1);}Poly g = f.eval(a);function<void(int, int, int)> dfs2 = [&](int p, int l, int r) {int m = l + r >> 1;if (l == r) {P[p].a.push_back(y[m] * power(g[m], mod - 2));return;}dfs2(p << 1, l, m), dfs2(p << 1 | 1, m + 1, r);P[p].a.resize(r - l + 1);Poly A = P[p << 1] * Q[p << 1 | 1];Poly B = P[p << 1 | 1] * Q[p << 1];for (int i = 0; i <= r - l; i ++) {P[p][i] = A[i] + B[i];}};dfs2(1, 0, a.size() - 1);return P[1];}
};
Poly toFPP(vector<Z> &a) {int n = a.size();vector<Z> b(n);iota(b.begin(), b.end(), 0);auto F = Poly(a).eval(b);vector<Z> f(n), g(n);for (int i = 0, sign = 1; i < n; i ++, sign *= -1) {f[i] = F[i] * infact[i];g[i] = Z(sign) * infact[i];}return Poly(f) * Poly(g);
}
Poly toOP(vector<Z> &a) {int n = a.size();vector<Z> g(n);for (int i = 0; i < n; i ++) {g[i] = infact[i];}auto F = Poly(a) * Poly(g);for (int i = 0; i < n; i ++) {F[i] *= fact[i];}vector<Z> p(n);iota(p.begin(), p.end(), 0);return Poly(p).inter(F);
}
Poly FPPMul(Poly a, Poly b) {int n = a.size() + b.size() - 1;Poly p;p.resize(n);for (int i = 0; i < n; i ++) {p[i] = infact[i];}a *= p, b *= p;for (int i = 0; i < n; i ++) {a[i] *= b[i] * fact[i];}for (int i = 1; i < n; i += 2) {p[i] = -p[i];}a *= p;a.resize(n);return a;
}
Poly Lagrange2(vector<Z> &f, int m, int k) {int n = f.size() - 1;vector<Z> a(n + 1), b(n + 1 + k);for (int i = 0; i <= n; i ++) {a[i] = f[i] * ((n - i) & 1 ? -1 : 1) * infact[n - i] * infact[i];}for (int i = 0; i <= n + k; i ++) {b[i] = Z(1) / (m - n + i);}Poly ans = Poly(a) * Poly(b);for (int i = 0; i <= k; i ++) {ans[i] = ans[i + n];}ans.resize(k + 1);Z sum = 1;for (int i = 0; i <= n; i ++) {sum *= m - i;}for (int i = 0; i <= k; i ++) {ans[i] *= sum;sum *= Z(m + i + 1) / (m - n + i);}return ans;
}
Poly Chirp_Z_Transform(vector<Z> &a, int c, int m) {int n = a.size();a.resize(n + m - 1);Poly f, g;f.resize(n + m - 1), g.resize(n + m - 1);for (int i = 0; i < n + m - 1; i ++) {f[n - 1 + m - 1 - i] = power(Z(c), i * (i - 1LL) / 2 % (mod - 1));g[i] = power(Z(c), mod - 1 - i * (i - 1LL) / 2 % (mod - 1)) * a[i];}Poly res = f * g, ans;ans.resize(m);for (int i = 0; i < m; i ++) {ans[i] = res[n - 1 + m - 1 - i] * power(Z(c), mod - 1 - i * (i - 1LL) / 2 % (mod - 1));}return ans;
}
Poly S2_row;
void S2_row_init(int n) {vector<Z> f(n + 1), g(n + 1);for (int i = 0; i <= n; i ++) {f[i] = power(Z(i), n) * infact[i];g[i] = Z(i & 1 ? -1 : 1) * infact[i];}S2_row = Poly(f) * Poly(g);
}
Poly S2_col;
void S2_col_init(int n, int k) {n ++;vector<Z> f(n);for (int i = 1; i < n; i ++) {f[i] = infact[i];}auto ans = Poly(f).pow(k, n);S2_col.resize(n + 1);for (int i = 0; i < n; i ++) {S2_col[i] = ans[i] * fact[i] * infact[k];}
}
Poly Bell;
void Bell_init(int n) {vector<Z> f(n + 1);for (int i = 1; i <= n; i ++) {f[i] = infact[i];}auto ans = Poly(f).exp(n + 1);Bell.resize(n + 1);for (int i = 0; i <= n; i ++) {Bell[i] = ans[i] * fact[i];}
}
vector<Z> p;
void p_init(int n) {vector<int> f(n + 1);p.resize(n + 1);p[0] = 1;f[0] = 1, f[1] = 2, f[2] = 5, f[3] = 7;for (int i = 4; f[i - 1] <= n; i ++) {f[i] = 3 + 2 * f[i - 2] - f[i - 4];}for (int i = 1; i <= n; i ++) {for (int j = 0; f[j] <= i; j ++) {p[i] += Z(j & 2 ? -1 : 1) * p[i - f[j]];}}
}
Poly P;
void p_init(int n, int m) {vector<Z> a(n + 1);for (int i = 1; i <= m; i ++) {for (int j = i; j <= n; j += i) {a[j] += Z(j / i).inv();}}P = Poly(a).exp(n + 1);
}
signed main() {cin.tie(0) -> sync_with_stdio(0);int n, m, k;cin >> n >> m >> k;int sz = min(n, k);Poly f;f.resize(sz + 1);f[0] = 2;Z fsum = 1, fn = n;for (int i = 1; i <= sz; i ++) {fsum = fsum * fn / i;f[i] = i & 1 ? Z(0) : fsum * 2;fn -= 1;}cout << f.pow(m, k + 1)[k] << "\n";
}