578_D. LCS Again_18717
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.Scanner;

public class Main {
	private void solve() {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int m = scanner.nextInt();
		String s = scanner.next();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = s.charAt(i) - 'a';
		}

		long ans = n * (m - 1);
		int[] same = new int[n];
		for (int i = 0; i < n - 1; i++) {
			if (i > 0 && a[i + 1] == a[i - 1]) {
				same[i + 1] = same[i - 1] + 1;
			}
			if (a[i] != a[i + 1]) {
				ans += (m - 1) * n;
				int x = same[i];
				int y = same[i + 1];
				int length = (1 + Math.min(x, y)) * 2;
				if (y > x) {
					length++;
				}
				ans -= length - 1;
			}
		}
		System.out.println(ans);
		scanner.close();
	}

	public static void main(String[] args) {
		new Main().solve();
	}
}
