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 MOD = 1000000007;
template <class T>
typename T::value_type arr_sum(const T &v, int n) {
  typename T::value_type sum = 0;
  for (int i = (0); i < (n); ++i) sum += v[i];
  return sum;
}
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
int lcs(const string &s1, const string &s2) {
  vector<vector<int>> dp(s1.size() + 1, vector<int>(s2.size() + 1));
  for (int i = (1); i < (s1.size() + 1); ++i) {
    for (int j = (1); j < (s2.size() + 1); ++j) {
      if (s1[i - 1] == s2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1] + 1;
      } else {
        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
      }
    }
  }
  return dp[s1.size()][s2.size()];
}
int main() {
  long long n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  long long t = 0;
  long long t2 = 0;
  for (int i = (1); i < (n); ++i) {
    if (s[i] == s[i - 1]) {
      ++t;
    }
  }
  for (int i = (1); i < (n); ++i) {
    long long cnt = 0;
    if (s[i] != s[i - 1]) {
      while (s[i + 1] == s[i - 1]) {
        ++i;
        ++cnt;
      }
    }
    t2 += cnt * (cnt + 1) / 2;
  }
  long long res = (long long)n * (n - t) * (m - 1) - (n - t - 1) - t2;
  cout << res;
}
