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<vector>
#include<cmath>
#include<map>
#include<cstdlib>
#include<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
#include<stack>
#include<bitset>
#include<functional>
#include<ctime>
#include<queue>
#include<deque>
#include<complex>
#include<cassert>
using namespace std;
#define pb push_back
#define pf push_front
typedef long long lint;
typedef complex<double> P;
#define mp make_pair
#define fi first
#define se second
typedef pair<int,int> pint;
#define All(s) s.begin(),s.end()
#define rAll(s) s.rbegin(),s.rend()
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)
//問題文および制約はちゃんと確認しよう！
//サイズは10^5じゃなくて2×10^5とかかもしれないし、重要な制約・条件を見落としているかも
//とりあえずサンプルを読んでから解法を考えよう？
int dp[510][510][510];
int num[510];
lint co[510][510];
lint mo=1000000007;
int main()
{
	rep(i,505){
		co[i][0]=co[i][i]=1;
		REP(j,1,i) co[i][j]=(co[i-1][j]+co[i-1][j-1])%mo;
	}
	memset(num,0,sizeof(num));
	int n,L,x;
	cin>>n>>x;
	rep(i,n){
		cin>>L;num[L]++;
	}
	memset(dp,0,sizeof(dp));
	dp[500][0][0]=1;
	for(int i=500;i>0;i--) rep(j,500/i+5) rep(k,505){
		if(dp[i][j][k]<1) continue;
		//cout<<i<<' '<<j<<' '<<k<<' '<<dp[i][j][k]<<endl;
		lint t=dp[i][j][k];
		lint can=j+k-(i+1)*j;
		if(j<1) can=0;
		rep(l,num[i+1]){
			t*=can;t%=mo;
		}
		/*if(i<1){
			dp[i][j][k]=(int)(t%mo);continue;
		}*/
		for(int l=0;k+i*l<=x;l++){
			dp[i-1][j+l][k+i*l]=(dp[i-1][j+l][k+i*l]+t*co[j+l][l])%mo;
			//dp[i-1][j+l][k+i*l]%=mo;
		}
	}
	lint out=0;
	rep(i,x+1) REP(j,1,x+1){
		if(dp[0][i][j]<1) continue;
		if(j+i-1>x) continue;
		//dp[0][i][j]=(dp[0][i][j]*num[1])%mo;
		//cout<<0<<' '<<i<<' '<<j<<' '<<dp[0][i][j]<<endl;
		lint t=dp[0][i][j];
		rep(k,num[1]){
			t*=j;t%=mo;
		}
		//cout<<0<<' '<<i<<' '<<j<<' '<<t<<endl;
		t*=co[x-j+1][i];t%=mo;
		if((x-j)%2==0) out+=t;else out-=t;
		out%=mo;
	}
	out%=mo;out+=mo;out%=mo;
	cout<<out<<endl;
}