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<bits/stdc++.h>

using namespace std;

#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);i--)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
#define Eunique(v) v.erace(unique(all(v)),v.end())
#define pvec(v)　rep(i,v.size()){cout << v[i] << " ";}
#define INF 0x3f3f3f3f
#define mset(m,v) memset(m,v,sezeof(m))
#define mod 1e9 +7;

const double pi = acos(-1.0);

typedef vector<int> vi;
typedef string ss;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef deque<ll> dll;
typedef pair<ll,ll> P;
ll dx[4] = {1,0,-1,0},dy[4] = {0,1,0,-1};

ll gcd(ll x,ll y)
{
    ll r;
    while((r = x % y) != 0)
    {
        x = y;
        y = r;
    }
    return y;
}

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

int main()
{
    ll n,k;
    cin >> n >> k;

    if(n == 1){
        cout << k << endl;
        return 0;
    }
    cout << (int)k*pow(k-1,n-1) << endl;
    return 0;
}