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>
#define rep(i,n)for(int i=0;i<n;++i)
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int, int>P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
//max=({});
//条件式が真ならwhileの中身を回し続ける
//printf("%d\n", ans);

//pairの入力
//vector<pair<ll, ll>>work(n);
//rep(i, n) {
//	ll a, b;
//	cin >> a >> b;
//	work[i] = make_pair(a, b);
//for(auto p:mp)(mapの探索)
//printf("%.10f\n",なんちゃら)

int g[15][15];
const int INF = 1001001001;
const int dx[4] = { -1,0,1,0 };
const int dy[4] = { 0,-1,0,1 };

//最大公約数
ll gcd(ll x, ll y) {
	return y ? gcd(y, x % y) : x;
}

ll lcm(ll x, ll y) {
	if (x == 0 || y == 0)return 0;
	return (x / gcd(x, y) * y);
}

//素因数分解
vector<pair<ll, int>>factorize(ll n) {
	vector<pair<ll, int>>res;
	for (ll i = 2;i * i <= n;++i) {
		if (n % i)continue;
		res.emplace_back(i, 0);
		while (n % i == 0) {
			n /= i;
			res.back().second++;
		}
	}
	if (n != 1)res.emplace_back(n, 1);
	return res;
}

int bingo[3][3];
int cnt[2][105];
int h, w;
bool flag[110][110];

int main() {
	ll n, k;
	cin >> n >> k;
	ll ans = k;
	rep(i, n-1) {
		ans *= (k - 1);
	}
	cout << ans << endl:
	return 0;
}
