Problem: There are n students in a class working on group projects. The students will divide into groups (some students may be in groups alone), work on their independent pieces, and then discuss the results together. It takes the i-th student ai minutes to finish his/her independent piece.

If students work at different paces, it can be frustrating for the faster students and stressful for the slower ones. In particular, the imbalance of a group is defined as the maximum ai in the group minus the minimum ai in the group. Note that a group containing a single student has an imbalance of 0. How many ways are there for the students to divide into groups so that the total imbalance of all groups is at most k?

Two divisions are considered distinct if there exists a pair of students who work in the same group in one division but different groups in the other.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 200, 0 ≤ k ≤ 1000) — the number of students and the maximum total imbalance allowed, respectively.

The second line contains n space-separated integers ai (1 ≤ ai ≤ 500) — the time it takes the i-th student to complete his/her independent piece of work.

Output

Print a single integer, the number of ways the students can form groups. As the answer may be large, print its value modulo 109 + 7.

Examples

Input

3 2
2 4 5


Output

3


Input

4 3
7 8 9 10


Output

13


Input

4 0
5 10 20 21


Output

1

Note

In the first sample, we have three options: 

  * The first and second students form a group, and the third student forms a group. Total imbalance is 2 + 0 = 2. 
  * The first student forms a group, and the second and third students form a group. Total imbalance is 0 + 1 = 1. 
  * All three students form their own groups. Total imbalance is 0. 



In the third sample, the total imbalance must be 0, so each student must work individually.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream &operator>>(istream &is, vector<pair<T1, T2>> &v) {
  for (pair<T1, T2> &t : v) is >> t.first >> t.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << " ";
  }
  os << '\n';
  return os;
}
double pi = acos(-1.0);
int md = 1e9 + 7;
const int INF = 1e15;
int dx1[] = {0, 0, -1, 1};
int dy1[] = {1, -1, 0, 0};
template <class T>
T abst(T a) {
  return a < 0 ? -a : a;
}
template <class T>
T max2(T a, T b) {
  return a > b ? a : b;
}
template <class T>
T min2(T a, T b) {
  return a < b ? a : b;
}
template <class T>
T maxm(initializer_list<T> l) {
  T ans = -INF;
  for (T i : l) ans = max2(ans, i);
  return ans;
}
int n, k;
vector<int> v;
int dp[201][201][1001];
int fun(int i, int j, int c) {
  if (c > k or j < 0) return 0;
  if (i == n) return j == 0;
  if (dp[i][j][c] != -1) return dp[i][j][c];
  long long int ans = 0;
  ans = ((1LL * j) * (1LL * fun(i + 1, j, c + j * (v[i] - v[i - 1])))) % md;
  ans = (ans +
         ((1LL * j) * (1LL * fun(i + 1, j - 1, c + j * (v[i] - v[i - 1])))) %
             md) %
        md;
  ans = (ans + 1LL * fun(i + 1, j + 1, c + j * (v[i] - v[i - 1]))) % md;
  ans = (ans + 1LL * fun(i + 1, j, c + j * (v[i] - v[i - 1]))) % md;
  return dp[i][j][c] = ans;
}
void solve() {
  cin >> n >> k;
  v.resize(n);
  cin >> v;
  sort((v).begin(), (v).end());
  memset(dp, -1, sizeof(dp));
  int ans = (fun(1, 1, 0) + fun(1, 0, 0)) % md;
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
}