There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors.

Find the number of the possible ways to paint the balls.

Constraints

* 1≦N≦1000
* 2≦K≦1000
* The correct answer is at most 2^{31}-1.

Input

The input is given from Standard Input in the following format:


N K


Output

Print the number of the possible ways to paint the balls.

Examples

Input

2 2


Output

2


Input

1 10


Output

10
#include<bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i=0, i##_len=(n); i<i##_len; ++i)
typedef long long ll;
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(),(x).end()
#define RALL(x) (x).rbegin(),(x).rend()
template <class T, class U> ostream &operator<<(ostream &os, const pair<T, U> &p){os<<"("<<p.first<<","<< p.second<<")";return os;}
template <class T> ostream &operator<<(ostream &os, const vector<T> &v){os<<"{";REP(i,(int)v.size()){if(i)os<<",";os<<v[i];}os<<"}";return os;}
template <typename T, size_t S> void printArray(const T (&array)[S]){for(auto val : array)std::cout << val << ", ";std::cout << "\n";}
void Yes() {cout << "Yes" << endl;}
void No() {cout << "No" << endl;}
void YES() {cout << "YES" << endl;}
void NO() {cout << "NO" << endl;}
const double PI=acos(-1);
const ll MOD = 1000000007;
using Graph = vector<vector<int>>;
//const int dx[4] = {1, 0, -1, 0};
//const int dy[4] = {0, 1, 0, -1};
//const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
//const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
ll mypow(ll a, ll n){
    ll res = 1;
    while(n>0){
        if(n&1) res = res*a;
        a = a*a;
        n >>= 1;
    }
    return res;
}
void solve(){
    ll n, k; cin >> n >> k;
    cout << k*mypow(k-1, n-1)) << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
}