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<iostream>
#include<algorithm>
#include<vector>
#include<string.h>
#include<cstdlib>
#include<map>
#include<fstream>
#include<string>
#include<sstream>
#include<iomanip>
#include<cmath>

using namespace std;

long long dp[3][100100] = {};
const long long INF = 1LL << 60;
#define MOD 1000000007

template<class T> inline bool chmin(T & a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

/**
* @fn
* 最大値を取ります.Vector<int>を渡すこと.
*/
long long int imax(vector<int> a) {
	int MAX = -10000000;
	for (int i = 0; i < a.size(); i++) {
		if (MAX < a.at(i)) MAX = a.at(i);
	}
	return MAX;
}

/**
* @fn
* 最小値を取ります.Vector<int>を渡すこと.
*/
long long int imin(vector<int> a) {
	int MIN = 1000000000;
	for (int i = 0; i < a.size(); i++) {
		if (MIN > a.at(i)) MIN = a.at(i);
	}
	return MIN;
}


long long int absmin(vector<int> a, long long int b) {
	int mini = 100000000;
	int d = 0;
	for (int i = 0; i < a.size(); i++) {
		if (mini > abs(a.at(i) - b)) {
			mini = abs(a.at(i) - b);
			d = i;
		}
	}
	return d;
}

int gcd(int a, int b)
{
	if (a < b) swap(a, b);
	if (b < 1) return -1;
	if (a % b == 0) return b;
	return gcd(b, a % b);
}

//ここまでライブラリ
//ここから本文

long long int tpimin(vector<int> a) {
	int MIN = 1000000000;
	int number = 0;
	for (int i = 0; i < a.size(); i++) {
		if (MIN > a.at(i)) MIN = a.at(i);
		number = i;
	}
	return number;
}


int main() {
	long long n, k;
	cin >> n >> k;
	cout << long long (k * pow(k - 1, n - 1)) << endl;
}