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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T, class U>
inline void print(T first, U second) {
  printsp(first);
  println(second);
}
template <class T, class U, class V>
inline void print(T first, U second, V z) {
  printsp(first);
  printsp(second);
  println(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
void add(int &a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
}
const int C = 2222;
int dp[C][C], ans, tmp = 1;
char s[C];
void solve_C() {
  int n, k;
  gn(n, k);
  scanf("%s", s + 1);
  dp[0][0] = 1;
  if (!k) {
    ans = 1;
    for (int i = 1; i <= n; i++) {
      dp[i][0] = (long long)tmp * (s[i] - 'a') % 1000000007;
      add(tmp, dp[i][0]);
      add(ans, dp[i][0]);
    }
    println(ans);
    return;
  }
  for (int i = 1; i <= n; i++) {
    dp[i][0] = (long long)tmp * (s[i] - 'a') % 1000000007;
    add(tmp, dp[i][0]);
  }
  for (int j = 1; j <= k; j++) {
    tmp = 0;
    for (int i = 1; i <= n; i++) {
      add(dp[i][j], (long long)tmp * (s[i] - 'a') % 1000000007);
      int flag = 0;
      for (int r = 1; j >= r * (n - i + 1) && r <= i; r++) {
        add(flag, (long long)dp[i - r][j - r * (n - i + 1)] * ('z' - s[i]) %
                      1000000007);
      }
      add(dp[i][j], flag);
      add(tmp, dp[i][j]);
      if (j == k) {
        add(ans, dp[i][j]);
      }
    }
  }
  println(ans);
  exit(0);
}
int main() {
  solve_C();
  return 0;
}
