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 N = 1e5 + 7;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-6;
const double PI = acos(-1.0);
char s[N];
int f[N];
long long calc(int n, int m) {
  int cnt = 0;
  long long Ans = 0;
  for (int i = n; i; --i) {
    if (i == n || s[i] != s[i + 1]) {
      ++cnt;
    }
    Ans += (cnt - 1LL) * (m - 1LL);
  }
  return Ans;
}
int main(void) {
  int n, m;
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  long long Ans = (m - 1LL) * n;
  Ans += calc(n, m);
  reverse(s + 1, s + n + 1);
  Ans += calc(n, m);
  for (int i = 1; i < n; ++i)
    if (s[i] != s[i + 1]) --Ans;
  for (int i = 2; i < n; ++i) {
    if (s[i - 1] == s[i + 1]) f[i + 1] = f[i] + 1;
    if (s[i - 1] == s[i + 1] && s[i - 1] != s[i]) Ans -= f[i + 1];
  }
  cout << Ans << endl;
  return 0;
}
