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;
using ll = long long;
using ld = long double;
const int p = 1000000007;
int mul(int a, int b) { return (1LL * a * b) % p; }
int add(int a, int b) {
  int s = (a + b);
  if (s >= p) s -= p;
  return s;
}
int sub(int a, int b) {
  int s = (a + p - b);
  if (s >= p) s -= p;
  return s;
}
int po(int a, int deg) {
  if (deg == 0) return 1;
  if (deg % 2 == 1) return mul(a, po(a, deg - 1));
  int t = po(a, deg / 2);
  return mul(t, t);
}
int inv(int n) { return po(n, p - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  ll res = n * (m - 1);
  if (n == 1) {
    cout << m - 1;
    return 0;
  }
  for (int i = 1; i < n; i++)
    if (s[i - 1] != s[i]) res += (m - 1) * i;
  for (int i = n - 2; i >= 0; i--)
    if (s[i] != s[i + 1]) res += (m - 1) * (n - 1 - i);
  int last = 0;
  for (int i = 1; i < n; i++) {
    if (i > 1) {
      if (s[i] != s[i - 2]) last = i - 1;
    }
    if (s[i] != s[i - 1]) res -= (i - last);
  }
  cout << res;
}
