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;
template <typename T>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
string to_string(char c) { return "'" + string(1, c) + "'"; }
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
template <typename A>
string to_string(A);
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ": " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool f = false;
  string r = "{";
  for (auto x : v) {
    if (f) r += ", ";
    r += to_string(x);
    f = true;
  }
  return r += "}";
}
template <typename A>
string to_string(vector<vector<A>> v) {
  string r;
  for (auto x : v) r += "\n" + to_string(x);
  return r;
}
void err(istream_iterator<string>) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " =: " << to_string(a) << "; ";
  err(++it, args...);
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
int const MOD = 998244353;
long long const INF = 1e18 + 42;
inline int add(int a, int b, int mod = MOD) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline int sub(int a, int b, int mod = MOD) {
  a -= b;
  return a < 0 ? a + mod : a;
}
inline int mul(int a, int b, int mod = MOD) {
  return int((long long)a * b % mod);
}
inline int mpow(int base, long long ex, int mod = MOD) {
  int res = 1;
  for (; ex > 0; ex >>= 1) {
    if (ex & 1) res = mul(res, base, mod);
    base = mul(base, base, mod);
  }
  return res;
}
inline int inv(int a, int mod = MOD) { return mpow(a, mod - 2, mod); }
inline int mdiv(int a, int b, int mod = MOD) {
  return mul(a, mpow(b, mod - 2, mod));
}
inline void adds(int &a, int b, int mod = MOD) {
  a += b;
  if (a >= mod) a -= mod;
}
inline void subs(int &a, int b, int mod = MOD) {
  a -= b;
  if (a < 0) a += mod;
}
inline void muls(int &a, int b, int mod = MOD) {
  a = int((long long)a * b % mod);
}
inline void mdivs(int &a, int b, int mod = MOD) { a = mdiv(a, b, mod); }
vector<int> fac, ifac;
inline int ncr(int n, int r) {
  if (n < r || r < 0 || n < 0) return 0;
  return mul(fac[n], mul(ifac[n - r], ifac[r]));
}
void prepare_factorial(int nax = 1e6 + 10) {
  fac.resize(nax);
  ifac.resize(nax);
  fac[0] = 1;
  for (int i = 1; i < nax; ++i) fac[i] = mul(i, fac[i - 1]);
  ifac[nax - 1] = inv(fac[nax - 1]);
  for (int i = nax - 1; i > 0; --i) ifac[i - 1] = mul(i, ifac[i]);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  prepare_factorial(5005);
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> y(k + 1);
  vector<int> p(k + 1, 1), q(k + 1, 1), pwk(k + 1);
  p[1] = inv(m), q[1] = sub(1, p[1]);
  pwk[1] = 1;
  for (int i = 2; i <= k; ++i) {
    p[i] = mul(p[i - 1], p[1]);
    q[i] = mul(q[i - 1], q[1]);
    pwk[i] = mpow(i, k);
  }
  for (int i = 0; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      adds(y[i], mul(pwk[j], mul(ncr(i, j), mul(p[j], q[i - j]))));
    }
  }
  if (n <= k) kek(y[n]);
  int ans = 0;
  for (int i = 0; i <= k; ++i) {
    int num = y[i], den = 1;
    for (int j = 0; j <= k; ++j)
      if (i != j) {
        muls(num, n - j);
        muls(den, (i - j + MOD) % MOD);
      }
    adds(ans, mdiv(num, den));
  }
  kek(ans);
}