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>
template <typename T>
inline bool MIN(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool MAX(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
inline int plu(int a, int b) {
  return a + b >= 1000000007 ? a + b - 1000000007 : a + b;
}
inline int sub(int a, int b) { return a - b < 0 ? a - b + 1000000007 : a - b; }
inline int mul(int a, int b) { return (long long)a * b % 1000000007; }
inline int inc(int &a, int b) { a = plu(a, b); }
using namespace std;
int n, m;
char s[100010];
int f[100010];
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  long long ans = n * (m - 1);
  for (int i = 1; i < n; ++i)
    if (s[i] != s[i + 1]) {
      ans += (m - 1) * n;
    }
  for (int i = 1; i <= n; ++i) {
    if (i == 1 || s[i] == s[i - 1]) {
      f[i] = 1;
    } else if (i - 2 > 0 && s[i] == s[i - 2]) {
      f[i] = f[i - 1] + 1;
    } else
      f[i] = 2;
    ans -= (f[i] - 1ll);
  }
  printf("%I64d\n", ans);
  return 0;
}
