Problem

There are $ M $ type characters. Use them to create a string of length $ N $. How many strings are used that have $ K $ or more? Find too much divided by $ 998244353 $.

Here, the difference between two strings of length $ N $ is defined as follows.

* If the two strings are $ S = S_1S_2 \ ldots S_N $, $ T = T_1T_2 \ ldots T_N $, then $ i $$ (1 \ leq i \ leq N) becomes $ S_i \ neq T_i $ ) $ Exists.

Constraints

The input satisfies the following conditions.

* $ 1 \ leq M \ leq 10 ^ {18} $
* $ 1 \ leq N \ leq 10 ^ {18} $
* $ 1 \ leq K \ leq 1000 $
* All inputs are integers

Input

The input is given in the following format.


$ M $ $ N $ $ K $


$ M, N, K $ are given on one line, separated by blanks.

Output

Output the remainder of the number of characters used in a string of length $ N $ that is greater than or equal to $ K $ divided by $ 998244353 $.

Examples

Input

2 10 1


Output

1024


Input

1 1 2


Output

0


Input

5 10 3


Output

9755400


Input

7 8 3


Output

5759460


Input

1000000000000000000 1000000000000000000 1000


Output

133611974
// #define _GLIBCXX_DEBUG // for STL debug (optional)
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;
using ll = long long int;
using int64 = long long int;
 
template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}
 
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
const int INF = 1001001001;
const ll LONGINF = 1001001001001001LL;
const ll MOD = 1000000007LL;

// ModInt begin

using ll = long long;
template<ll mod>
struct ModInt {
    ll v;
    ll mod_pow(ll x, ll n) const {
        return (!n) ? 1 : (mod_pow((x*x)%mod,n/2) * ((n&1)?x:1)) % mod;
    }
    ModInt(ll a = 0) : v((a %= mod) < 0 ? a + mod : a) {}
    ModInt operator+ ( const ModInt& b ) const {
        return (v + b.v >= mod ? ModInt(v + b.v - mod) : ModInt(v + b.v));
    }
    ModInt operator- () const {
        return ModInt(-v);
    }
    ModInt operator- ( const ModInt& b ) const {
        return (v - b.v < 0 ? ModInt(v - b.v + mod) : ModInt(v - b.v));
    }
    ModInt operator* ( const ModInt& b ) const {return (v * b.v) % mod;}
    ModInt operator/ ( const ModInt& b ) const {return (v * mod_pow(b.v, mod-2)) % mod;}
    
    bool operator== ( const ModInt &b ) const {return v == b.v;}
    bool operator!= ( const ModInt &b ) const {return !(*this == b); }
    ModInt& operator+= ( const ModInt &b ) {
        v += b.v;
        if(v >= mod) v -= mod;
        return *this;
    }
    ModInt& operator-= ( const ModInt &b ) {
        v -= b.v;
        if(v < 0) v += mod;
        return *this;
    }
    ModInt& operator*= ( const ModInt &b ) {
        (v *= b.v) %= mod;
        return *this;
    }
    ModInt& operator/= ( const ModInt &b ) {
        (v *= mod_pow(b.v, mod-2)) %= mod;
        return *this;
    }
    ModInt pow(ll x) { return ModInt(mod_pow(v, x)); }
    // operator int() const { return int(v); }
    // operator long long int() const { return v; }
};

template<ll mod>
ostream& operator<< (ostream& out, ModInt<mod> a) {return out << a.v;}
template<ll mod>
istream& operator>> (istream& in, ModInt<mod>& a) {
    in >> a.v;
    return in;
}

// ModInt end

using mint = ModInt<998244353>;

mint comb(ll N, ll K) {
    if(K < 0 or N < K) return mint(0);
    mint res(1);
    if(K > N - K) K = N - K;
    for(ll i=0; i<K; i++) {
        res *= mint(N - i);
        res /= mint(i + 1);
    }
    return res;
}

const int MAXN = 1005;
mint pas[MAXN+5][MAXN+5];
mint powtab[MAXN+5];

int main() {
    ll M, N, K; cin >> M >> N >> K;

    for(int i=0; i<=MAXN; i++) {
        powtab[i] = mint(i).pow(N);
        pas[i][0] = 1;
        for(int j=1; j<=i; j++) pas[i][j] = pas[i-1][j-1] + pas[i-1][j];
    }
    
    mint ans = mint(M).pow(N);
    for(int x=1; x<K; x++) {
        mint add = comb(M, x);
        mint num = 0;
        for(int k=1; k<=x; k++) {
            if((x - k) % 2 == 1) num -= powtab[k] * pas[x][k];
            else num += powtab[k] * pas[x][k];
        }
        ans -= add * num;
        // cerr << "x = " << x << ", add = " << add << ", num = " << num << endl;
    }
    cout << ans << endl;
    return 0;
}

