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.
import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
    public static void main (String[] args) throws java.lang.Exception
    {
        // your code goes here
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        String s = sc.next();
        long ans = n*m-n;
        for(int i = 1;i<n;i++){
            if(s.charAt(i)!=s.charAt(i-1)){
                ans+=n*m-n;
            }
        }
        long cur = 1;
        for(int i = 1;i<n;i++){
            if(cur==1){
                if(s.charAt(i)==s.charAt(i-1)){
                    continue;
                }
                else{
                    cur++;
                }
            }
            else{
                if(s.charAt(i)==s.charAt(i-2)){
                    cur++;
                }
                else{
                    ans-=(cur*(cur-1))/2;;
                    if(s.charAt(i)==s.charAt(i-1)){
                        cur=1;
                    }
                    else{
                        cur=2;
                    }
                }
            }
        }
        ans-=(cur*(cur-1))/2;
        System.out.println(ans);
    }
}