Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

2 2

Sample Output 1

2

We will denote the colors by 0 and 1. There are two possible ways: we can either paint the left ball in color 0 and the right ball in color 1, or paint the left in color 1 and the right in color 0.

Sample Input 2

1 10

Sample Output 2

10

Since there is only one ball, we can use any of the ten colors to paint it. Thus, the answer is ten.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<algorithm>
#include<numeric>
#include<cmath>
#include<queue>
#include<stack>
#include<list>
#include<unordered_map>
#include<unordered_set>
#include<cctype>
#include<iomanip>

using namespace std;
typedef long long int ll;typedef pair<int, int> P;
const long long mod = 1e9 + 7;
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define all(a) (a).begin(),(a).end()
#define allg(a) (a).begin(),(a).end(),greater<>()
#define SUM(a) accumulate((a).begin(),(a).end(),0)
const double PI = 3.14159265359;

int main(void) {
	ll n, k; cin >> n >> k;
	ll ans = k;
	k--;
	for (ll i = 0; i < n; i++) {
		ans *= k;
	}
	cout << ans << endl;
	return 0;
}