//
// Created by whq on 2021/2/21.
//

#include <bits/stdc++.h>

#define uset unordered_set
#define umap unordered_map
#define debug cout<<"ok"<<endl
#define ll long long
#define ii pair<ll, ll>
#define fi first
#define se second
#define vi vector<int>
#define vb vector<bool>
#define vvi vector<vector<int>>
#define vvb vector<vector<bool>>
#define vvl vector<vector<ll>>
#define vl vector<ll>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define For(i, n) for (int i = 0; i < n; i++)
#define forn(i, a, b) for(int i=a;i<=b;i++)
#define pb(a) push_back(a)
#define all(a) a.begin(),a.end()
#define lowbit(x) x&(-x)
using namespace std;

class A {
public:
    void solve() {
        ll k;
        cin >> k;
        ll res = 0;
        for (ll i = 1; i * i * i <= k; i++) res++;
        for (ll i = 1; i * (i + 1) <= k; i++) {
            for (ll j = i + 1; j * i <= k; j++) {
                if (i * i * j <= k) res += 3;
                if (i * j * j <= k) res += 3;
            }
        }
        for (ll i = 1; i * (i + 1) <= k; i++) {
            for (ll j = i + 1; j * i <= k; j++) {
                for (ll l = j + 1; l * j * i <= k; l++) res += 6;
            }
        }
        cout << res << endl;
    }
};

class B {
public:
    ll base;

    ll fpow(ll a, ll k) {
        if (k == 0) return 1;
        ll temp = fpow(a, k / 2) % base;
        temp = (temp * temp) % base;
        if ((k % 2) == 1) return ((temp % base) * (a % base)) % base;
        else return temp;
    }

    ll superPow(ll a, ll b) {
        if(b==0) return 1ll;
        ll p1=fpow(a,b%10);
        ll p2=fpow(superPow(a,b/10),10);
        return (p1)*(p2)%base;
    }
    void solve() {
        ll a, b, c;
        cin >> a >> b >> c;
        set<int> one = {0, 1, 5, 6};
        ll cur = a % 10;
        map<int, map<int, int> > two;
        two[4][1] = 4;
        two[4][0] = 6;
        two[9][1] = 9;
        two[9][0] = 1;
        if (one.count(cur)) {
            cout << cur << endl;
            return;
        }
        if (cur == 4 || cur == 9) {
            cout << two[cur][b % 2] << endl;
            return;
        }
        base = 4;
        map<int, map<int, int>> f;
        f[2][1] = 2;
        f[2][2] = 4;
        f[2][3] = 8;
        f[2][0] = 6;
        f[3][1] = 3;
        f[3][2] = 9;
        f[3][3] = 7;
        f[3][0] = 1;
        f[7][1] = 7;
        f[7][2] = 9;
        f[7][3] = 3;
        f[7][0] = 1;
        f[8][1] = 8;
        f[8][2] = 4;
        f[8][3] = 2;
        f[8][0] = 6;
        ll left=superPow(b,c)%base;
        cout<<f[cur][left]<<endl;
    }
};
class C{
public:
    void solve(){
        string s;
        cin>>s;
        vi cnt(26,0);
        ll n=s.size();
        ll res=0;
        for(int i=n-1;i>=0;i--){
            cnt[s[i]-'a']++;
            if(i+1<n&&s[i]==s[i+1]){
                res+=n-i-cnt[s[i]-'a'];
                for(int j=0;j<26;j++){
                    if(j==s[i]-'a') continue;
                    cnt[s[i]-'a']+=cnt[j];
                    cnt[j]=0;
                }
            }
        }
        cout<<res<<endl;
    }
};
class D{
public:
    ll base=998244353;
    ll binpow(ll a, ll b) {
        ll res = 1;
        while (b > 0) {
            if (b & 1) res = res * a%base;
            a = a * a%base;
            b >>= 1;
        }
        return res;
    }
    void solve(){
        ll n,m,k;
        cin>>n>>m>>k;
        ll res=0;
        if(n==1||m==1){
            cout<<binpow(k,n*m)<<endl;
            return;
        }
        for(ll i=1;i<=k;i++) {
            ll tot=(binpow(i,n)-binpow(i-1,n))%mod;
            res=(res+tot*(binpow(k-i+1,m))%base)%base;
        }
        if(res<0) res+=base;
        cout<<res%base<<endl;
    }
};
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    D ans;
    ans.solve();
    return 0;
}
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <vector>
using namespace std;
using ll = long long;
#define rep(i, j, n) for (ll i = j; i < (n); ++i)
#define rrep(i, j, n) for (ll i = (n)-1; j <= i; --i)

template <typename T>
std::ostream& operator<<(std::ostream& os, std::vector<T>& a) {
  for (size_t i = 0; i < a.size(); ++i) os << (i > 0 ? " " : "") << a[i];
  return os << '\n';
}

[[maybe_unused]] constexpr ll MOD = 998244353;
[[maybe_unused]] constexpr int INF = 0x3f3f3f3f;
[[maybe_unused]] constexpr ll INFL = 0x3f3f3f3f3f3f3f3fLL;

ll mpow(ll x, ll e) {
  ll res = 1;
  while (e > 0) {
    if (e & 1) res = res * x % MOD;
    x = x * x % MOD;
    e >>= 1;
  }
  return res;
}

int main() {
  cin.tie(0)->sync_with_stdio(0);
  ll n, m, k;
  cin >> n >> m >> k;
  if (n == 1 || m == 1) {
    if (n == 1 && m == 1) {
      cout << k;
      return 0;
    }
    if (n == 1) {
      ll ans = 0;
      rep(i, 1, k + 1) {
        // 行の最小値がi
        // 少なくとも人つ
        ans += (MOD + (mpow(k - (i - 1), m) - mpow(k - i, m)) % MOD) % MOD;
        ans %= MOD;
      }
      cout << ans;
      return 0;
    }
    if (m == 1) {
      ll ans = 0;
      rep(i, 1, k + 1) {
        ans += (MOD + (mpow(i, n) - mpow(i - 1, n)) % MOD) % MOD;
        ans %= MOD;
      }
      cout << ans;
      return 0;
    }
  }

  // aのみの問題を考える。　➡　各行は独立なので、普通にk^n
  // マスの埋め方は
  // a[i]==xの時、i行目にはx未満の値が現れない　∧　xが少なくとも一回現れる。

  // k^nで決めた後、aのmax以上の値は、任意のbで取れる。
  // aのmaxがxとなる組み合わせ * (x ~ k)^m
  ll ans = 0;
  rep(x, 1, k + 1) {
    ll tmp = (MOD + (mpow(x, n) - mpow(x - 1, n)) % MOD) % MOD;
    ans += tmp * mpow(k - x + 1, m) % MOD;
    ans %= MOD;
  }
  cout << ans;

  return 0;
}