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.io.*;
import java.util.Scanner;


public class Main1 {
	static int hashi(int x, int y)
	{
		if (x == y) return -1;
		if (x > y) {
			int temp = x;
			x = y;
			y = temp;
		}
		return x * 1000 + y;
	}
	public static void main(String[] args) {
		
		Scanner sc = new Scanner(System.in);
		int n, m;
		int id[] = new int[100005];
		
		 long ans = 0;
		String s;//[100005];
		
		n = sc.nextInt();
		m = sc.nextInt();
		String str = sc.next();
		s = " " + str;
		
			ans = (m - 1) * n;
			for (int i = 2; i <= n; i++)
				if (s.charAt(i - 1) != s.charAt(i))
					ans +=  (i - 1) * (m - 1);
			
			for (int i = 1; i <= n - 1; i++)
				if (s.charAt(i) != s.charAt(i + 1))
					ans +=(n - i) * (m - 1);
			
			for (int i = 1; i <= n - 1; i++) 
				id[i] = hashi(s.charAt(i), s.charAt(i + 1));
			
			for (int i = 1; i <= n - 1; i++){
				if (id[i] != -1 && (i == 1 || id[i] != id[i - 1])){
					for (int j = i; j <= n - 1 && id[j] == id[i]; j++){
						ans -= j - i + 1;
					}
						
				}
					
			}
				
			
			System.out.println(ans);
		}
	}

