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>
struct MI {
 private:
  char bb[4096];
  FILE* f;
  char *bs, *be;
  char e;
  bool o, l;

 public:
  MI() : f(stdin) {}
  inline char get() {
    if (o) {
      o = 0;
      return e;
    }
    if (bs == be) be = (bs = bb) + fread(bb, 1, sizeof(bb), f);
    if (bs == be) {
      l = 1;
      return -1;
    };
    return *bs++;
  }
  inline void unget(char c) {
    o = 1;
    e = c;
  }
  template <typename T>
  inline MI& operator>(T&);
};
template <typename T>
struct Q {
  const static bool U = T(-1) >= T(0);
  inline void operator()(MI& t, T& r) const {
    r = 0;
    char c;
    bool y = 0;
    if (U)
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (isdigit(c)) break;
      }
    else
      for (;;) {
        c = t.get();
        if (c == -1) goto E;
        if (c == '-') {
          c = t.get();
          if (isdigit(c)) {
            y = 1;
            break;
          };
        } else if (isdigit(c))
          break;
        ;
      };
    for (;;) {
      if (c == -1) goto E;
      if (isdigit(c))
        r = r * 10 + (c ^ 48);
      else
        break;
      c = t.get();
    }
    t.unget(c);
  E:;
    if (y) r = -r;
  }
};
template <>
struct Q<char> {};
template <typename T>
inline MI& MI::operator>(T& t) {
  Q<T>()(*this, t);
  return *this;
}
template <typename T>
std::ostream& operator<(std::ostream& out, const T& t) {
  return out << t;
}
using std::cout;
MI cin;
const int kmax = 5005;
const int mod = 998244353;
inline long long ksm(long long x, int p = mod - 2) {
  long long ret = 1;
  while (p) {
    if (p & 1) (ret *= x) %= mod;
    (x *= x) %= mod;
    p >>= 1;
  }
  return ret;
}
inline int pro(int x) { return x >= mod ? x - mod : x; }
int n, m, K, S[kmax][kmax], fac[kmax], ifac[kmax];
inline void init(int n) {
  for (int i = fac[0] = ifac[0] = 1; i <= K; ++i)
    fac[i] = (long long)fac[i - 1] * i % mod;
  ifac[n] = ksm(fac[n]);
  for (int i = K; i > 1; --i) ifac[i - 1] = (long long)ifac[i] * i % mod;
}
int main() {
  cin > n > m > K;
  m = ksm(m);
  init(K);
  S[0][0] = 1;
  for (int i = 1; i <= K; ++i)
    for (int j = 1; j <= i; ++j)
      S[i][j] = pro(S[i - 1][j - 1] + (long long)S[i - 1][j] * j % mod);
  int ret = 0;
  for (int i = 0, cur = 1; i <= K; ++i) {
    ret = pro(ret + (long long)S[K][i] * cur % mod);
    cur = (long long)cur * (n - i) % mod * m % mod;
  }
  cout < ret < ('\n');
}