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;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do { (t*=10)+=ch-'0'; ch=getchar(); } while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const ll mod=(1e9)+7;
int n,L,X,cnt[510],C[510][510];
ll dp[2][510][510],tmp,ans,ivjc[510],iv[510],jc[510];
ll ksm(ll x,ll y) {
	ll res=1;
	while (y) {
		if (y&1) res=res*x%mod;
		x=x*x%mod; y>>=1;
	} return res;
}
void update(ll &x,ll y) {
	x+=y; if (x>=mod) x-=mod;
}
int main() {
	//freopen("1.txt","r",stdin);
	read(n); read(X);
	for (int i=1;i<=n;i++) read(L),cnt[L]++;
	for (int i=0;i<=X;i++) {
		C[i][0]=C[i][i]=1;
		for (int j=1;j<i;j++) C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
	}
	iv[0]=jc[0]=ivjc[0]=1;
	iv[1]=jc[1]=ivjc[1]=1;
	for (int i=2;i<=X;i++) {
		jc[i]=jc[i-1]*i%mod;
		iv[i]=(mod-mod/i)*iv[mod%i]%mod;
		ivjc[i]=ivjc[i-1]*iv[i]%mod;
	}
	dp[X&1^1][0][0]=1;
	for (int i=X;i>=1;i--) {
		for (int j=0;j<=X;j++)
		for (int k=i*j;k<=X;k++) {
			dp[i&1][j][k]=0;
			tmp=ksm(k-(i-1)*j,cnt[i]);
			for (int l=0;l<=min(j,k/i);l++)
				update(dp[i&1][j][k],dp[i&1^1][j-l][k-i*l]*C[j][l]%mod*tmp%mod);
		}
	}
	for (int i=0;i<=X;i++)
	for (int j=0;j<=min(i+1,X);j++) {
		tmp=C[i+1][j]*dp[1][j][X-i]%mod;
		if (i&1) tmp=mod-tmp;
		update(ans,tmp);
	}
	printf("%lld\n",ans);
	return 0;
}
/*
  0. Enough array size? Enough array size? Enough array size? Interger overflow?
  
  1. Think TWICE, Code ONCE!
  Are there any counterexamples to your algo?
    
  2. Be careful about the BOUNDARIES!
  N=1? P=1? Something about 0?
    
  3. Do not make STUPID MISTAKES!
  Time complexity? Memory usage? Precision error?
*/