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
#include <bits/stdc++.h>
using namespace std;
int dp[2000 + 1][2000 + 1];
int sp[2000 + 1];
char str[2000 + 10];
char *trim_line(char *str) { return strtok(str, "\r\n"); }
int addmod(int x, int y, int p) {
  long long sum = x;
  sum += y;
  if (sum >= p) sum -= p;
  return (int)sum;
}
int submod(int x, int y, int p) {
  long long diff = x;
  diff -= y;
  if (diff < 0) diff += p;
  return (int)diff;
}
int mulmod(int x, int y, int p) {
  long long prod = x;
  prod = (prod * y) % p;
  return (int)prod;
}
int muladdmod(int x, int y, int z, int p) {
  long long prod = x;
  prod = (prod * y + z) % p;
  return (int)prod;
}
int solve_problem() {
  char *p;
  int n, k;
  if (fgets(str, sizeof(str), stdin) == NULL) return 1;
  p = strtok(str, " ");
  n = atoi(p);
  p = strtok(NULL, " ");
  k = atoi(p);
  if (fgets(str, sizeof(str), stdin) == NULL) return 1;
  sp[0] = dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= k; j++) {
      char ch = str[i - 1];
      int step = n + 1 - i;
      int delta = 0;
      dp[i][j] = muladdmod(sp[j], ch - 'a', dp[i][j], 1000000007);
      for (int r = 0, q = j - step; r < i && q >= 0; r++, q -= step)
        delta = addmod(delta, dp[i - 1 - r][q], 1000000007);
      dp[i][j] = muladdmod(delta, 'z' - ch, dp[i][j], 1000000007);
      sp[j] = addmod(sp[j], dp[i][j], 1000000007);
    }
  int result = 0;
  for (int i = 0; i <= n; i++) result = addmod(result, dp[i][k], 1000000007);
  printf("%d\n", result);
  return 0;
}
int main() {
  solve_problem();
  return 0;
}
