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;
using ll = long long;
using ld = long double;
const int MOD = 1e9 + 7;
const int INF = 1e18 + 9;
constexpr long double pi = 3.141592653589793238462643383279;
int fact(int i) {
  if (i == 0) return 1;
  return ((fact(i - 1)) * i) % MOD;
}
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
int keta(int n) {
  if (n == 0) return 1;
  int count = 0;
  while (n != 0) {
    n /= 10;
    count++;
  }
  return count;
}
int ketasum(int n) {
  int sum = 0;
  while (n != 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
char toSmall(char c) { return (c + 0x20); }
char toLarge(char c) { return (c - 0x20); }
float myPower(int a, int n) {
  float x = 1;
  while (n > 0) {
    if (n & 1) {
      x *= a;
    }
    a *= a;
    n >>= 1;
  }
  return x;
}
bool is_prime(const unsigned n) {
  switch (n) {
    case 0:
    case 1:
      return false;
    case 2:
    case 3:
      return true;
  }
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (unsigned i = 5; i * i <= n; i += 6) {
    if (n % i == 0) return false;
    if (n % (i + 2) == 0) return false;
  }
  return true;
}
int main(void) {
  int n, k;
  cin >> n >> k;
  cout << k * pow(k - 1, n - 1) << endl;
  return 0;
}
