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: Accepted)

#include <bits/stdc++.h>
using namespace std;

// macros
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define MIN(a,b) (a>b?b:a)
#define MAX(a,b) (a>b?a:b)
#define INT_IN(a) int a; cin >> a
#define STR_IN(s) string s; cin >> s
#define PRINTLN(s) cout << s << endl
#define PRINTDOUBLELN(s) cout << setprecision(12) << (s) << endl;

// util functions
template<class T> inline T sqr(T x) {return x*x;}

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

template <typename T>inline T max(vector<T> v){T max=v[0]; FOR(i, 1, v.size())max=MAX(v[i], max); return max;}
template <typename T>inline T min(vector<T> v){T min=v[0]; FOR(i, 1, v.size())min=MIN(v[i], min); return min;}
template <typename T>inline vector<T> concat(vector<T> a, vector<T> b){vector<T> ab; ab.reserve(a.size()+b.size());ab.insert(ab.end(),a.begin(),a.end());ab.insert(ab.end(),b.begin(),b.end());return ab;}

// types
typedef long long LL;

// consts
const double EPS = 1e-10;
const double PI  = acos(-1.0);

//debug
#define dump(x)  #x << " = " << (x) << endl
#define debug(x) #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl

int main(){
    INT_IN(N);
    INT_IN(K);
    int p = K;
    REP(i, N - 1){
        p *= K - 1;
    }
    PRINTLN(p);
    return 0;
}