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 MAXN = 1e6 + 100;
const int inf_int = 1e9 + 100;
const long long inf_ll = 1e18;
const double PI = 3.1415926535898;
bool debug = 0;
const int MOD = 1e9 + 7;
int n, k;
int less1[MAXN];
int bigg1[MAXN];
inline int add(int a, int b) {
  a = a + b;
  if (a >= MOD) a -= MOD;
  return a;
}
inline int mul(int a, int b) { return (1ll * a * b) % MOD; }
int dp[2020][2020];
int pref[2020][2020];
void solve() {
  cin >> n >> k;
  string s;
  cin >> s;
  s = "#" + s;
  for (int i = 1; i < (int)s.size(); ++i) {
    less1[i] = s[i] - 'a';
    bigg1[i] = 'z' - s[i];
  }
  dp[0][0] = 1;
  pref[0][0] = 1;
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int e = 0; e <= k; ++e) {
      dp[i][e] = mul(add(dp[i][e], pref[i - 1][e]), less1[i]);
      if (debug) cerr << "  $ " << i << " " << e << " " << dp[i][e] << endl;
      for (int eq = 0; e - 1ll * (1 + eq) * (n - i + 1) >= 0 && i - eq - 1 >= 0;
           eq++) {
        if (debug)
          cerr << "  $ "
               << "try " << i - eq - 1 << " " << e - (1 + eq) * (n - i + 1)
               << " " << endl;
        dp[i][e] = add(dp[i][e], mul(dp[i - eq - 1][e - (1 + eq) * (n - i + 1)],
                                     bigg1[i]));
      }
      if (debug) cerr << "  $ " << i << " " << e << " " << dp[i][e] << endl;
      pref[i][e] = add(pref[i - 1][e], dp[i][e]);
    }
  }
  if (k == 0) ans++;
  for (int i = 1; i <= n; ++i) {
    ans = add(ans, dp[i][k]);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(15);
  solve();
}
