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;
typedef long long ll;
typedef long long LL;
typedef long double lb;
#define ri register int
const lb PI=3.141592653589793238462643383279;
const ll inf=1000000000000000000ll;
const int N=100005,M=1000000007;
int n,i,j,k,l,x,m;
long long f[505][505][505],a[505],inv[505],ans,fac[505];
long long qpow(long long a,long long b)
{
	long long s=a,ans=1;
	while(b)
	{
		if(b&1)
			ans=ans*s%M;
		s=s*s%M;
		b>>=1;
	}
	return ans;
}
int main()
{
	scanf("%d %d",&n,&x);
	for(i=1;i<=n;++i)
	{
		scanf("%d",&m);
		++a[m];
	}
	fac[0]=inv[0]=1;
	for(i=1;i<=x;++i)
	{
		inv[i]=inv[i-1]*qpow(i,M-2)%M;
		fac[i]=fac[i-1]*i%M;
	}
	f[x+1][0][0]=1;
	for(i=x;i>=1;--i)
	{
		for(j=0;j<=x;++j)
			for(k=0;x-k+1>=j;++k)
				for(l=0;l*i<=k&&l<=j;++l)
					f[i][j][k]=(f[i][j][k]+f[i+1][j-l][k-l*i]*inv[l])%M;
		if(a[i])
		{
			for(j=0;j<=x;++j)
				for(k=0;k<=x;++k)
					f[i][j][k]=f[i][j][k]*qpow(k-(i-1)*j,a[i])%M;
		}
	}
	for(i=0;i<=x;++i)
		for(j=0;j<=x;++j)
			if(x-j+1>=i)
				ans=(ans+f[1][i][j]*fac[x-j+1]%M*inv[x-j+1-i]*((x-j)&1?-1:1))%M;
	cout<<(ans%M+M)%M;
}