Levko loves strings of length n, consisting of lowercase English letters, very much. He has one such string s. For each string t of length n, Levko defines its beauty relative to s as the number of pairs of indexes i, j (1 ≤ i ≤ j ≤ n), such that substring t[i..j] is lexicographically larger than substring s[i..j].

The boy wondered how many strings t are there, such that their beauty relative to s equals exactly k. Help him, find the remainder after division this number by 1000000007 (109 + 7).

A substring s[i..j] of string s = s1s2... sn is string sisi + 1... sj.

String x = x1x2... xp is lexicographically larger than string y = y1y2... yp, if there is such number r (r < p), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1. The string characters are compared by their ASCII codes.

Input

The first line contains two integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 2000).

The second line contains a non-empty string s of length n. String s consists only of lowercase English letters. 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
yz


Output

26


Input

2 3
yx


Output

2


Input

4 7
abcd


Output

21962
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class LevcoNStrings {
    static int[][][] dp;
    static char[] str;
    static int n;
    static TreeSet<String> S = new TreeSet<>();
    static long MOD = 1000000007;

    public static int f(int i, int k, int s) {
        if (i == str.length)
            return k == 0 ? 1 : 0;
        if (dp[i][k][s] != -1) {
            return dp[i][k][s];
        }
        long result = (str[i] - 'a' + 0L) * f(i + 1, k, 0) + f(i + 1, k, 1);
        if (s == 0) {
            int upto = 0;
            for (int j = i; j < str.length && (j - i + 1) * (n - j) <= k; j++) {
                result += f(j + 1, k - (j - i + 1) * (n - j), 0)
                        * ('z' + 0L - str[j]);
                upto = j;
            }
            for (int j = str.length - 1; j > upto && (j - i + 1) * (n - j) <= k; j--) {
                result += f(j + 1, k - (j - i + 1) * (n - j), 0)
                        * ('z' + 0L - str[j]);
            }
        }
        result %= MOD;
        return dp[i][k][s] = (int) result;
    }

    public static void main(String[] args) throws Exception {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer strtok;
        strtok = new StringTokenizer(in.readLine());
        n = Integer.parseInt(strtok.nextToken());
        int k = Integer.parseInt(strtok.nextToken());

        strtok = new StringTokenizer(in.readLine());
        str = strtok.nextToken().toCharArray();

        dp = new int[n + 1][k + 1][2];
        for (int[][] a : dp)
            for (int[] b : a)
                Arrays.fill(b, -1);
        System.out.println(f(0, k, 0));
    }
}
