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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author sheep
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskC {
    private static final int MOD = 1000000007;
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int n, k;
        n = in.nextInt();
        k = in.nextInt();
        char[] s = in.next().toCharArray();

        long[][] dp = new long[n + 1][k + 1];
        long[][] sum = new long[n + 1][k + 1];
        dp[0][0] = sum[0][0] = 1;
        long result = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j <= k; ++j) {
                dp[i][j] = (dp[i][j] + sum[i - 1][j] * (s[i - 1] - 'a')) % MOD;
                for (int t = 0; (n - i + 1) * (t + 1) <= j && i - 1 - t >= 0; ++t) {
                    dp[i][j] = (dp[i][j] + dp[i - 1 - t][j - (n - i + 1) * (t + 1)] * ('z' - s[i - 1])) % MOD;
                }
                sum[i][j] = sum[i - 1][j] + dp[i][j];
            }
        }
        for (int i = 0; i <= n; ++i) {
            result += dp[i][k];
        }
        out.println(result % MOD);
    }
}

class InputReader {
    BufferedReader reader;
    StringTokenizer tokenizer;

    public InputReader(InputStream stream) {
        reader = new BufferedReader(new InputStreamReader(stream));
        tokenizer = null;
    }

    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (Exception e) {
                throw new UnknownError();
            }
        }
        return tokenizer.nextToken();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    }
