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.

Examples

Input

2 2


Output

2


Input

1 10


Output

10
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool amax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool amin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
int LCM(int a, int b) { return a * b / GCD(a, b); }
long long LCM(long long a, long long b) { return a * b / GCD(a, b); }
struct aaa {
  aaa() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} aaaaaaa;
const int INF = 1001001001;
const long long LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const int dx[] = {1, 1, 0, -1, -1, -1, 0, 1},
          dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (long long i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}
long long COM(long long n, long long k) {
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
signed main() {
  long long n, k;
  cin >> n >> k;
  COMinit();
  cout << COM(n, k) << endl;
  return 0;
}
