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;
int T, N, A, w[100100], G, CNT;
long long ans;
string s;
int main() {
  scanf("%d %d", &N, &A);
  map<int, int> m;
  cin >> s;
  for (int i = (0); i < (N); i++) w[i] = s[i] - 'a';
  for (int i = (0); i < (N); i++) {
    scanf("%d", &w[i]);
    w[i] = (m.count(w[i]) ? m[w[i]] : m[w[i]] = ++CNT);
  }
  long long add = 0;
  for (int i = (0); i < (N); i++)
    if (!i || w[i - 1] != w[i]) {
      G++;
      add += -N + 1;
    }
  ans = 1ll * G * N * A;
  ans = (ans + add);
  for (int i = (1); i < (N); i++) {
    int a = w[i - 1], b = w[i];
    if (a == b) continue;
    if (i - 2 < 0 || w[i - 2] != b) {
      int j = i + 1;
      while (j < N) {
        int n = (j - i) % 2 ? a : b;
        if (w[j] == n)
          j++;
        else
          break;
      }
      long long x = 1ll * (j - i + 1) * (j - i + 2) / 2 - 1;
      ans -= x;
    }
  }
  printf("%lld\n", ans - 1);
  return 0;
}
