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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C360 {
    
    static StringTokenizer st;
    static BufferedReader in;
    static PrintWriter pw;
    
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n = nextInt();
        int k = nextInt();
        String s = " "+next();
        long[][]dp = new long[n+1][k+1];
        long[]sum = new long[k+1];
        int mod = (int) (1e9+7);
        dp[0][0] = 1;
        sum[0] = 1;
        for (int i = 1; i <= n; i++) {
            int w = s.charAt(i);
            for (int j = 0; j <= k; j++) {
                dp[i][j] = sum[j] * (w-'a') % mod;
                for (int j2 = 0; j2 < i; j2++) {
                    int t = (j2+1) * (n-i+1);
                    if (t > j)
                        break;
                    dp[i][j] = (dp[i][j] + dp[i-j2-1][j-t] * ('z'-w)) % mod;
                }
                sum[j] = (sum[j] + dp[i][j]) % mod;
            }
        }
        System.out.println(sum[k]);
        pw.close();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }
    
    private static double nextDouble() throws IOException{
        return Double.parseDouble(next());
    }
    
    private static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
}
