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, class U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (int i = 0, i_len = ((int)v.size()); i < i_len; ++i) {
    if (i) os << ",";
    os << v[i];
  }
  os << "}";
  return os;
}
template <typename T, size_t S>
void printArray(const T (&array)[S]) {
  for (auto val : array) std::cout << val << ", ";
  std::cout << "\n";
}
void Yes() { cout << "Yes" << endl; }
void No() { cout << "No" << endl; }
void YES() { cout << "YES" << endl; }
void NO() { cout << "NO" << endl; }
const double PI = acos(-1);
const long long MOD = 1000000007;
using Graph = vector<vector<int>>;
long long mypow(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a;
    a = a * a;
    n >>= 1;
  }
  return res;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  cout << k * pow(k - 1, n - 1) << endl;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}
