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;
const int oo = 2139063143;
const int N = 2010;
const int P = 1000000007;
const double eps = 1e-7;
template <typename T>
inline void sc(T &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
template <typename T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void pr(T x) {
  print(x), putchar('\n');
}
template <typename T1, typename T2>
void chkmax(T1 &A, T2 B) {
  A = A >= B ? A : B;
}
template <typename T1, typename T2>
void chkmin(T1 &A, T2 B) {
  A = A <= B ? A : B;
}
int del(int x) { return x >= P ? x - P : x; }
void add(int &x, int y) { x = del(x + y); }
int n, m, f[N][N], g[N], a[N], b[N];
char s[N];
int main() {
  sc(n), sc(m);
  scanf("%s", s + 1);
  for (int i = 1; i <= n; i++) a[i] = 'z' - s[i], b[i] = 26 - a[i] - 1;
  f[0][0] = g[0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m; j++) {
      f[i][j] = (long long)b[i] * g[j] % P;
      int tmp = 0;
      for (int k = i - 1; k >= 0; k--) {
        int p = j - (i - k) * (n - i + 1);
        if (p < 0) break;
        add(tmp, f[k][p]);
      }
      add(f[i][j], (long long)tmp * a[i] % P);
      add(g[j], f[i][j]);
    }
  pr(g[m]);
  return 0;
}
