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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define REP(i, n)   for(int i = 0; i < (n); i++)
#define REPS(i, n)  for(int i = 1; i <= (n); i++)
#define RREP(i, n)  for(int i = (n)-1; i >= 0; i--)
#define RREPS(i, n) for(int i = (n); i > 0; i--)
#define ALL(v) v.begin(), v.end()
#define RALL(v) v.rbegin(), v.rend()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define mp make_pair
#define mt make_tuple
#define pb push_back

using ll  = long long;
using pi  = pair<int, int>;
using pl  = pair<ll, ll>;
using vi  = vector<int>;
using vl  = vector<ll>;
using vs  = vector<string>;
using vb  = vector<bool>;
using vvi = vector<vi>;
using vvl = vector<vl>;

const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const ll  INFL = 1e18;
const double PI = 3.141592653589793;
const double EPS = 1e-9;

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

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);

    int N, K; cin >> N >> K;
    cout << K * (int)pow(K-1, N-1) << endl;
}