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<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
const int mod=1000000007;
inline int ad(int x){return x>=mod?x-mod:x;}
inline int dec(int x){return x<0?x+mod:x;}
inline int power(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)ans=1LL*ans*a%mod;
		a=1LL*a*a%mod;
		b>>=1;
	}return ans;
}
int cnt[510],f[510][510][510];
int fac[510],inv[510];
inline int C(int n,int m){return 1LL*fac[n]*inv[m]%mod*inv[n-m]%mod;}
int main()
{
	//freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
	fac[0]=1;
	for(int i=1;i<=500;i++)fac[i]=1LL*fac[i-1]*i%mod;inv[500]=power(fac[500],mod-2);
	for(int i=500;i>=1;i--)inv[i-1]=1LL*inv[i]*i%mod;
	int n=read(),m=read();
	for(int i=1;i<=n;i++)cnt[read()]++;
	f[m+1][0][0]=1;
	for(int i=m;i>=1;i--)
	{
		for(int j=0;j<=m;j++)
		{
			for(int k=0;k-1+j<=m;k++)
			{
				if(!f[i+1][j][k])continue;
				for(int u=0;u*i+j+u+k-1<=m;u++)
				{
					int w1=u*i+j,w2=u+k;
					f[i][w1][w2]=ad(f[i][w1][w2]+1LL*f[i+1][j][k]*power(w1-w2*(i-1),cnt[i])%mod*inv[u]%mod);
				}
			}
		}
	}
	int ans=0;
	for(int i=0;i<=m;i++)
	{
		int sum=0;
		for(int j=1;j-1+i<=m;j++)sum=ad(sum+1LL*f[1][i][j]*C(m-i+1,j)%mod*fac[j]%mod);
		if((m-i)&1)ans=dec(ans-sum);
		else ans=ad(ans+sum);
	}
	pr2(ans);
	return 0;
}