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>
#pragma GCC optimize("Ofast,unroll-loops")
using namespace std;
using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int INT_INF = (int)(2e9);
const ll LL_INF = (ll)(2e18);
const int NIL = -1;
static mt19937 _g(time(nullptr));
inline ll randint(ll a, ll b) {
  ll w = (_g() << 31LL) ^ _g();
  return a + w % (b - a + 1);
}
inline void fast_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
};
template <typename T>
inline T sign(T x) {
  return T(x > 0) - T(x < 0);
}
template <typename T, typename S>
inline ostream& operator<<(ostream& os, const pair<T, S> p) {
  cout << "[" << p.first << ";" << p.second << "]";
  return os;
}
template <typename T>
inline ostream& operator<<(ostream& os, const vector<T>& v) {
  for (auto el : v) cout << el << " ";
  return os;
}
template <typename T>
inline T fetch() {
  T ret;
  cin >> ret;
  return ret;
}
template <typename T>
inline vector<T> fetch_vec(int sz) {
  vector<T> ret(sz);
  for (auto& elem : ret) cin >> elem;
  return ret;
}
const int LOG = 31;
const int MAXK = 5077;
const int MOD = 998244353;
int add(int x, int y) { return (x + y >= MOD ? x + y - MOD : x + y); }
int sub(int x, int y) { return (x - y < 0 ? x - y + MOD : x - y); }
int mult(int x, int y) { return (x * 1LL * y) % MOD; }
int binpow(int x, int y) {
  if (y == 0 || x == 1) return 1;
  if (x == 0) return 0;
  int rez = binpow(x, y >> 1);
  rez = mult(rez, rez);
  if (y & 1) rez = mult(rez, x);
  return rez;
}
int divide(int x, int y) { return mult(x, binpow(y, MOD - 2)); }
const ll MOD2 = MOD * 1LL * MOD;
int pw[LOG][MAXK], rez[MAXK], temp[MAXK];
ll g[MAXK];
void solve() {
  int n = fetch<int>();
  int m = fetch<int>();
  int k = fetch<int>();
  vector<int> F(k + 1, 1), A(k + 1, 0);
  A[0] = 1;
  for (int i = 1; i <= k; ++i) {
    F[i] = mult(F[i - 1], i);
  }
  for (int i = 1; i <= k; ++i) {
    A[i] = divide(1, F[i]);
    A[i] = divide(A[i], m);
  }
  for (int i = 0; i <= k; ++i) {
    pw[0][i] = A[i];
  }
  rez[0] = 1;
  for (int log = 0; log < LOG - 1; ++log) {
    memset(g, 0, sizeof(g));
    for (int i = 0; i < MAXK; ++i) {
      for (int j = 0; i + j < MAXK; ++j) {
        g[i + j] += pw[log][i] * 1LL * pw[log][j];
        if (g[i + j] >= MOD2) {
          g[i + j] -= MOD2;
        }
      }
    }
    for (int i = 0; i < MAXK; ++i) {
      pw[log + 1][i] = g[i] % MOD;
    }
  }
  for (int log = 0; log < LOG; ++log) {
    if ((n >> log) & 1) {
      memset(g, 0, sizeof(g));
      for (int i = 0; i < MAXK; ++i) {
        for (int j = 0; i + j < MAXK; ++j) {
          g[i + j] += pw[log][i] * 1LL * rez[j];
          if (g[i + j] >= MOD2) {
            g[i + j] -= MOD2;
          }
        }
      }
      for (int i = 0; i < MAXK; ++i) {
        rez[i] = g[i] % MOD;
      }
    }
  }
  int answ = rez[k];
  answ = mult(answ, F[k]);
  cout << answ << '\n';
}
int main() {
  fast_io();
  solve();
  return 0;
}