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 <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int pct(int x) { return __builtin_popcount(x); }
int bits(int x) { return 31 - __builtin_clz(x); }
int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }
int fstTrue(function<bool(int)> first, int lo, int hi) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const long long INFLL = 0x3f3f3f3f3f3f3f3f;
const long long NINFLL = 0xc0c0c0c0c0c0c0c0;
const long long MOD = 1e9 + 7;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1},
          dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
long long modInv(long long a) { return binpow(a, MOD - 2); }
bool sortcol(const vector<long long>& v1, const vector<long long>& v2) {
  return v1[1] < v2[1];
}
bool sortpair(const pair<int, int>& p1, const pair<int, int>& p2) {
  return p1.first < p2.first;
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<long long unsigned> distribution(0, 10);
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int first = 0;
  cerr << '{';
  for (auto& i : x) cerr << (first++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int mxN = 2e5 + 5;
int q = 1, n, m, arr[505];
long long dp[2][205][1005], ans;
void solve() {
  cin >> n >> m;
  for (int i = (0); i < (n); i++) cin >> arr[i];
  sort(arr, arr + n);
  dp[0][0][0] = dp[0][1][0] = 1;
  for (int i = (0); i < (n); i++) {
    for (int j = (0); j < (i + 1); j++) {
      for (int k = (0); k < (m + 1); k++) {
        int nk = k + j * (arr[i] - (i ? arr[i - 1] : 0));
        if (nk > m) continue;
        dp[1][j + 1][nk] = (dp[1][j + 1][nk] + dp[0][j][k]) % MOD;
        dp[1][j][nk] = (dp[1][j][nk] + dp[0][j][k]) % MOD;
        dp[1][j][nk] = (dp[1][j][nk] + j * dp[0][j][k]) % MOD;
        if (j) dp[1][j - 1][nk] = (dp[1][j - 1][nk] + j * dp[0][j][k]) % MOD;
      }
    }
    for (int j = (0); j < (205); j++)
      for (int k = (0); k < (1005); k++) {
        dp[0][j][k] = dp[1][j][k];
        dp[1][j][k] = 0;
      }
  }
  for (int k = (0); k < (m + 1); k++) ans = (ans + dp[0][0][k]) % MOD;
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  while (q--) {
    solve();
  }
}