You are given a string S of length n with each character being one of the first m lowercase English letters. 

Calculate how many different strings T of length n composed from the first m lowercase English letters exist such that the length of LCS (longest common subsequence) between S and T is n - 1.

Recall that LCS of two strings S and T is the longest string C such that C both in S and T as a subsequence.

Input

The first line contains two numbers n and m denoting the length of string S and number of first English lowercase characters forming the character set for strings (1 ≤ n ≤ 100 000, 2 ≤ m ≤ 26).

The second line contains string S.

Output

Print the only line containing the answer.

Examples

Input

3 3
aaa


Output

6


Input

3 3
aab


Output

11


Input

1 2
a


Output

1


Input

10 9
abacadefgh


Output

789

Note

For the first sample, the 6 possible strings T are: aab, aac, aba, aca, baa, caa. 

For the second sample, the 11 possible strings T are: aaa, aac, aba, abb, abc, aca, acb, baa, bab, caa, cab.

For the third sample, the only possible string T is b.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int maxm = 30;
int n, m, t[maxn];
long long dp[maxn][2][2][2];
char s[maxn];
long long DP(int x, int l1, int l2, int l3) {
  if (x > n) return l2 == n - 1;
  if (l1 < x - 3 || l2 < x - 2 || l3 < x - 2) return 0;
  long long &r = dp[x][l1 & 1][l2 & 1][l3 & 1];
  if (r != -1) return r;
  r = 0;
  for (int i = 1; i <= m; i++) {
    int nowl1 = (i == t[x - 1] ? l1 + 1 : l2);
    int nowl2 = (i == t[x] ? l2 + 1 : max(nowl1, l3));
    int nowl3 = (i == t[x + 1] ? l3 + 1 : nowl2);
    r += DP(x + 1, nowl1, nowl2, nowl3);
  }
  return r;
}
int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  for (int i = 1; i <= n; i++) t[i] = s[i] - 'a' + 1;
  memset(dp, -1, sizeof(dp));
  printf("%I64d\n", DP(1, 0, 0, 0));
  return 0;
}
