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;
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 100100;
long long dp[maxn][30];
char str[maxn];
int a[maxn];
int nex[22], now[22];
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", str);
  for (int i = 0; i <= n - 1; i++) a[i + 1] = str[i] - 'a' + 1;
  for (int i = 1; i <= m; i++) {
    int state = 1;
    if (a[1] == i) state |= 2;
    if (a[1] == i || a[2] == i) state |= 4;
    dp[1][state]++;
  }
  for (int i = 2; i <= n; i++)
    for (int state = 0; state <= 7; state++) {
      now[0] = i - 3 + (((state)&1) > 0);
      now[1] = i - 2 + (((state)&2) > 0);
      now[2] = i - 2 + (((state)&4) > 0);
      for (int j = 1; j <= m; j++) {
        for (int l = 1; l <= 3; l++) {
          if (a[i - 2 + l] == j)
            nex[l] = now[l - 1] + 1;
          else
            nex[l] = max(nex[l - 1], now[l]);
        }
        if (nex[1] >= i - 2 && nex[2] >= i - 1 && nex[3] >= i - 1)
          dp[i][(nex[1] - i + 2) | ((nex[2] - i + 1) << 1) |
                ((nex[3] - i + 1) << 2)] += dp[i - 1][state];
      }
    }
  long long ans = dp[n][0] + dp[n][1] + dp[n][4] + dp[n][5];
  cout << ans << endl;
}
