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;
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
long long mod = 998244353;
const int MAX_FAC = 1e5 + 1;
long long fac[MAX_FAC], facinv[MAX_FAC];
const int PRIME_RANGE = 1e5 + 1;
vector<long long> prime;
const int PF_RANGE = 1e5 + 1;
vector<long long> pf[PF_RANGE];
const int DIV_RANGE = 1e5 + 1;
vector<long long> divs[DIV_RANGE];
long long power(long long a, long long b, long long p = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % p;
    a = (a * a) % p;
    b /= 2;
  }
  return res;
}
bool flag_fac = 0;
void fill_fac() {
  flag_fac = 1;
  fac[0] = 1;
  facinv[0] = 1;
  for (long long i = 1; i < MAX_FAC; i++) {
    fac[i] = (i * fac[i - 1]) % mod;
    facinv[i] = power(fac[i], mod - 2, mod);
  }
}
long long ncr(long long n, long long r) {
  if (!flag_fac) fill_fac();
  if (r > n || r < 0 || n < 0) return 0;
  long long res;
  if (n < MAX_FAC) {
    res = (fac[n] * facinv[r]) % mod;
    res = (res * facinv[n - r]) % mod;
  } else {
    res = facinv[r];
    for (long long i = 0; i < r; i++) res = (res * (n - i)) % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
long long gcd(vector<T> a) {
  if (!a.size()) return 1;
  long long res = a[0];
  for (int i = 1; i < a.size(); i++) res = gcd(res, (long long)a[i]);
  return res;
}
void fill_prime() {
  vector<bool> vis(PRIME_RANGE, 0);
  for (long long i = 2; i < PRIME_RANGE; i++) {
    if (!vis[i])
      prime.push_back(i);
    else
      continue;
    for (long long j = i; j < PRIME_RANGE; j += i) {
      vis[j] = 1;
    }
  }
}
void fill_pf() {
  vector<bool> vis(PF_RANGE, 0);
  pf[1].push_back(1);
  for (long long i = 2; i < PF_RANGE; i++) {
    if (vis[i]) continue;
    for (long long j = i; j < PF_RANGE; j += i) {
      vis[j] = 1;
      long long temp = j;
      while (temp % i == 0) {
        pf[j].push_back(i);
        temp /= i;
      }
    }
  }
}
void fill_divs() {
  for (long long i = 1; i < DIV_RANGE; i++) {
    for (long long j = i; j < DIV_RANGE; j += i) divs[j].push_back(i);
  }
}
vector<int> factorise(int n) {
  vector<int> res;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) res.push_back(i);
    while (n % i == 0) n /= i;
  }
  if (n > 1) res.push_back(n);
  return res;
}
long long dp[5001][5001];
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  while (T--) {
    long long n, m, k;
    cin >> n >> m >> k;
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (long long i = 1; i <= k; i++)
      for (long long j = 1; j <= k; j++) {
        long long t1 = (dp[i - 1][j - 1] * (n - j + 1)) % mod;
        long long t2 = (dp[i - 1][j] * (j)) % mod;
        dp[i][j] = (t1 + t2) % mod;
      }
    long long ans = 0;
    for (long long i = 1; i <= k; i++) {
      long long t1 = (power(m, i * (mod - 2), mod) * dp[k][i]);
      ans = (ans + t1) % mod;
    }
    cout << ans << '\n';
  }
  return 0;
}