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: Wrong Answer)

#include<bits/stdc++.h>

#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define ll long long
#define pii pair<int,int>
#define all(x) begin(x), end(x)
#define loop(i,n) for(int i=0; i<n; i++)
#define rep(i,a,b,c) for(int i=a; i<b; i+=c)
#define brep(i,a,b,c) for(int i=a; i>b; i-=c)
#define tc(t) int t; cin>>t; while(t--)
#define sz(v) int((v).size())
#define pb push_back


using namespace std;

void __print(int x) {cout << x;}
void __print(long x) {cout << x;}
void __print(long long x) {std::cout << x;}
void __print(unsigned x) {std::cout << x;}
void __print(unsigned long x) {std::cout << x;}
void __print(unsigned long long x) {std::cout << x;}
void __print(float x) {std::cout << x;}
void __print(double x) {std::cout << x;}
void __print(long double x) {std::cout << x;}
void __print(char x) {cout << '\'' << x << '\'';}
void __print(const char *x) {cout << '\"' << x << '\"';}
void __print(const string &x) {cout << '\"' << x << '\"';}
void __print(bool x) {cout << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cout << '{'; __print(x.first); cout << ','; __print(x.second); cout << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cout << '{'; for (auto &i : x) cout << (f++ ? "," : ""), __print(i); cout << "}";}
void _print() {cout << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cout << ", "; _print(v...);}
template <typename T> inline void inp(T &any) {cin >> any;} template <typename T, typename... U>inline void inp(T &a, U &...b) {cin >> a; inp(b...);}
#define rin(args...) ll args; inp(args);
#ifndef ONLINE_JUDGE
#define debug(x...) cout << "[" << #x << "] = ["; _print(x)
#define matdeb(mat,row,col) cout << #mat << "::" << endl; printmat(mat,row,col); printpartition;
#define debugv(v,n) cout << #v << ":" << endl; printv(v,0,n-1); printpartition;
#else
#define debug(x...)
#define matdeb(mat,row,col)
#define debugv(v,n)
#endif

const int maxm = 1000000000 + 7;

const ll MOD = 1e9 + 7;
const int INF = 55;
ll dp [INF] [INF * INF];

double startTime;
double getCurrentTime() {
    return ((double)clock() - startTime) / CLOCKS_PER_SEC;
}


//----------->CODE BEGINS --------------->


int main()
{

#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    IOS;

    int n, k;

    cin >> n >> k;

    ll ans = 1;



    for (int i = 0; i < n; i++)
    {
        // cout<<ans<<endl;
        ans *= (k - i);
    }

    cout << ans << endl;



}