Problem: Consider the following experiment. You have a deck of m cards, and exactly one card is a joker. n times, you do the following: shuffle the deck, take the top card of the deck, look at it and return it into the deck.

Let x be the number of times you have taken the joker out of the deck during this experiment. Assuming that every time you shuffle the deck, all m! possible permutations of cards are equiprobable, what is the expected value of x^k? Print the answer modulo 998244353.

Input

The only line contains three integers n, m and k (1 ≤ n, m < 998244353, 1 ≤ k ≤ 5000).

Output

Print one integer — the expected value of x^k, taken modulo 998244353 (the answer can always be represented as an irreducible fraction a/b, where b mod 998244353 ≠ 0; you have to print a ⋅ b^{-1} mod 998244353).

Examples

Input


1 1 1


Output


1


Input


1 1 5000


Output


1


Input


2 2 2


Output


499122178


Input


998244352 1337 5000


Output


326459680

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, G = 3, Gi = 332748118;
int n, m, m2, invm2, k;
int fact[100010], inv[100010];
int nxt[100010], mi[100010], ss[100010];
int rev[270000], tot, bit;
long long A[270000], B[270000];
long long ans;
inline int read() {
  int x = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -x : x;
}
long long ksm(long long B, long long K, long long A = 1) {
  for (; K; K >>= 1, B = B * B % mod)
    if (K & 1) A = A * B % mod;
  return A;
}
void xxs() {
  nxt[0] = nxt[1] = 1;
  mi[1] = 1;
  mi[0] = 0;
  for (int i = 2; i <= 100000; i++) {
    if (!nxt[i]) mi[ss[++ss[0]] = nxt[i] = i] = ksm(i, k);
    for (int j = 1; i * ss[j] <= 100000; j++) {
      nxt[i * ss[j]] = ss[j];
      mi[i * ss[j]] = 1ll * mi[i] * mi[ss[j]] % mod;
      if (i % ss[j] == 0) break;
    }
  }
}
void prepare() {
  fact[0] = 1;
  for (int i = 1; i <= 100000; i++) fact[i] = 1ll * fact[i - 1] * i % mod;
  inv[100000] = ksm(fact[100000], mod - 2);
  for (int i = 100000 - 1; i >= 0; i--)
    inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
}
long long C(long long N, long long M) {
  if (N < M || M < 0) return 0;
  return fact[N] * inv[M] % mod * inv[N - M] % mod;
}
void solve() {
  m2 = mod + 1 - m;
  invm2 = ksm(m2, mod - 2);
  long long mim = 1, mim2 = ksm(m2, n);
  for (int i = 0; i <= n; i++) {
    (ans += C(n, i) * mim % mod * mim2 % mod * mi[i] % mod) %= mod;
    mim = mim * m % mod;
    mim2 = mim2 * invm2 % mod;
  }
  cout << ans << '\n';
}
void NTT(long long C[], int inv) {
  for (int i = 0; i < tot; i++)
    if (i < rev[i]) swap(C[i], C[rev[i]]);
  for (int mid = 1; mid < tot; mid <<= 1) {
    long long w1 = ksm(inv == 1 ? G : Gi, (mod - 1) / (mid << 1));
    for (int i = 0; i < tot; i += (mid << 1)) {
      long long wk = 1;
      for (int j = 0; j < mid; j++, wk = wk * w1 % mod) {
        long long x = C[i + j], y = wk * C[i + j + mid] % mod;
        C[i + j] = (x + y) % mod;
        C[i + j + mid] = (x - y + mod) % mod;
      }
    }
  }
  if (inv == 1) return;
  inv = ksm(tot, mod - 2);
  for (int i = 0; i < tot; i++) C[i] = C[i] * inv % mod;
}
int main() {
  n = read();
  m = ksm(read(), mod - 2);
  k = read();
  if (n == 1) return printf("1\n"), 0;
  prepare();
  xxs();
  if (n <= k) return solve(), 0;
  for (int i = 0; i <= k; i++) A[i] = ksm(i, k) * inv[i] % mod;
  for (int i = 0; i <= k; i++)
    if (i & 1)
      B[i] = -inv[i];
    else
      B[i] = inv[i];
  while ((1 << bit) <= k + k) bit++;
  tot = 1 << bit;
  for (int i = 1; i < tot; i++)
    rev[i] = (rev[i >> 1] >> 1 | ((i & 1) << (bit - 1)));
  NTT(A, 1);
  NTT(B, 1);
  for (int i = 0; i < tot; i++) A[i] = A[i] * B[i] % mod;
  NTT(A, -1);
  long long p = m, xjm = 1;
  ans = A[0];
  for (int i = 1; i <= k; i++) {
    (xjm *= n - i + 1) %= mod;
    (ans += A[i] % mod * p % mod * xjm % mod) %= mod;
    p = p * m % mod;
  }
  cout << (ans + mod) % mod << '\n';
}