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 string NAME = "360C";
const int INF = 1e9 + 7;
const int N = 2e3 + 8, B = 1e9 + 7;
int n, K;
long long f[N][N], sum[N];
string s;
void Enter() {
  cin >> n >> K;
  cin >> s;
  s = ' ' + s;
}
inline void Add(long long &a, const long long b) { a = (a + b) % B; }
void Process() {
  f[0][0] = sum[0] = 1;
  for (int i = (1), _b = (n); i <= _b; ++i)
    for (int j = (0), _b = (K); j <= _b; ++j) {
      Add(f[i][j], sum[j] * (s[i] - 'a'));
      for (int pref = 0; i - pref - 1 >= 0 && j - (n - i + 1) * (pref + 1) >= 0;
           ++pref) {
        Add(f[i][j],
            f[i - pref - 1][j - (pref + 1) * (n - i + 1)] * ('z' - s[i]));
      }
      Add(sum[j], f[i][j]);
    }
}
void Result() {
  long long res = 0;
  if (K == 0) res = 1;
  for (int i = (1), _b = (n); i <= _b; ++i) Add(res, f[i][K]);
  cout << (res + B) % B;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  Enter();
  Process();
  Result();
}
