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;
long long gcd(long long a, long long b) {
  long long r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100010;
int n, m;
char s[maxn];
int c[maxn];
long long f[maxn][2][2];
long long cal(int pos, int a, int b) {
  if (pos == n) return a == 0;
  if (f[pos][a][b] != -1) return f[pos][a][b];
  long long res = cal(pos + 1, a, b);
  if (a) {
    if (pos == n - 1)
      res += b == 0 ? m : 0;
    else
      res += (m - 1) * cal(pos + 2, a - 1, b);
  }
  if (b) {
    if (pos + a < n && s[pos + a - 1] != s[pos + a])
      res += cal(pos + 1, a, b - 1);
  }
  if (a && b) {
    if (pos < n - 1 && s[pos] != s[pos + 1])
      res += (m - 2) * cal(pos + 2, a - 1, b - 1);
    else if (pos == n - 1)
      res += m - 1;
  }
  return f[pos][a][b] = res;
}
void solve() {
  memset(f, -1, sizeof(f));
  scanf("%d%d%s", &n, &m, s);
  long long sum = 0;
  c[0] = 1;
  for (int i = (1); i < (n); i++) {
    if (s[i] != s[i - 1])
      c[i] = 2;
    else
      c[i] = 1;
    if (s[i] != s[i - 1] && s[i] == s[i - 2]) c[i] = c[i - 1] + 1;
    sum += c[i] - 1;
  }
  long long ans = cal(0, 1, 1) - sum;
  cout << ans;
}
int main() {
  solve();
  return 0;
}
