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 readInt() {
  int a;
  scanf("%d", &a);
  return a;
}
long long readLong() {
  long long a;
  scanf("%lld", &a);
  return a;
}
void readString(char *s) { scanf(" %s", s); }
const int N = 100005;
char s[N];
vector<pair<int, int>> comp;
int f[26];
int dp[N];
void solve() {
  int n = readInt();
  int m = readInt();
  readString(s + 1);
  int last = 1;
  for (int i = (1); i <= (n); ++i) {
    if (s[i] != s[i + 1]) {
      comp.emplace_back(s[i] - 'a', i - last + 1);
      last = i + 1;
    }
    f[s[i] - 'a']++;
  }
  long long ans = (int)(comp).size() * (n * 1ll * m - n);
  dp[1] = 1;
  for (int i = (2); i <= (n); ++i) {
    if (s[i] != s[i - 1] && s[i] == s[i - 2])
      dp[i] = 1 + dp[i - 1];
    else if (s[i] != s[i - 1])
      dp[i] = 2;
    else
      dp[i] = 1;
    ans -= dp[i] - 1;
  }
  printf("%lld", ans);
}
int main() {
  int t = 1;
  while (t--) solve();
  return 0;
}
