Problem: Niwango bought a piece of land that can be represented as a half-open interval [0, X).

Niwango will lay out N vinyl sheets on this land. The sheets are numbered 1,2, \ldots, N, and they are distinguishable. For Sheet i, he can choose an integer j such that 0 \leq j \leq X - L_i and cover [j, j + L_i) with this sheet.

Find the number of ways to cover the land with the sheets such that no point in [0, X) remains uncovered, modulo (10^9+7). We consider two ways to cover the land different if and only if there is an integer i (1 \leq i \leq N) such that the region covered by Sheet i is different.

Constraints

* 1 \leq N \leq 100
* 1 \leq L_i \leq X \leq 500
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N X
L_1 L_2 \ldots L_N


Output

Print the answer.

Examples

Input

3 3
1 1 2


Output

10


Input

18 477
324 31 27 227 9 21 41 29 50 34 2 362 92 11 13 17 183 119


Output

134796357

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

#define N 505

int n, X;
int L[N];
int dp[2][N][N];

const int mod = 1e9 + 7;

void add(int &x, int y) {
	x += y;
	if (x >= mod) x -= mod;
}
int mul(int x, int y) {return 1ll * x * y % mod;}

int main() {
	cin >> n >> X;
    for (int i = 1; i <= n; i ++) cin >> L[i];
    sort(L+1, L+n+1, greater <int> ());
    int u = 0, v = 1;
    dp[u][L[1]][1] = 1;
    for (int i = 2; i <= n; i ++) {
        int l = L[i];
        int mx = X / L[i-1];
        memset(dp[v], 0, sizeof dp[v]);
        for (int len = 1; len <= X; len ++) {
            for (int k = 1; k <= mx; k ++) if (dp[u][len][k]) {
                if (len + l <= X) add(dp[v][len+l][k+1], mul(dp[u][len][k], k+1));
                add(dp[v][len][k], mul(len-k*(l-1), dp[u][len][k]));
                for (int t = 1; t < l; t ++) if (len + t <= X) add(dp[v][len+t][k], mul(2*k, dp[u][len][k]));
                for (int t = 0; t < l-1; t ++) if (len + t <= X) add(dp[v][len+t][k-1], mul((k-1)*(l-t-1), dp[u][len][k]));
            }
        }
        swap(u, v);
    }
    int ans = 0;
    for (int i = 1; i <= n; i ++) add(ans, dp[u][X][i]);
    cout << ans << endl;
	return 0;
}