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 Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;}
  Mint operator-(Mint a) const{return Mint(v)-=a;}
  Mint operator*(Mint a) const{return Mint(v)*=a;}
  Mint operator/(Mint a) const{return Mint(v)/=a;}

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static Mint comb(long long n,int k){
    Mint num(1),dom(1);
    for(int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}

//INSERT ABOVE HERE
using M = Mint<int>;
const int MAX = 505;
M dp[MAX][MAX]={};
M nx[MAX][MAX]={};
signed main(){
  int n,x;
  cin>>n>>x;
  vector<int> ls(n);
  for(int i=0;i<n;i++) cin>>ls[i];
  sort(ls.rbegin(),ls.rend());

  dp[0][0]=1;
  for(int l:ls){
    // i component
    for(int i=0;i<=n;i++){
      // length j
      for(int j=0;j<=x;j++){
        if(dp[i][j]==M(0)) continue;
        assert(j>=i*l);

        // solo
        if(j+l<=x)
          nx[i+1][j+l]+=dp[i][j]*M(i+1);

        if(!(i>=1)) continue;

        // whole
        nx[i+0][j+0]+=dp[i][j]*M(j-i*(l-1));

        // cover one
        for(int k=1;k<l and j+k<=x;k++)
          nx[i+0][j+k]+=dp[i][j]*M(i)*M(2);

        if(!(i>=2)) continue;

        // cover two
        for(int k=0;k+2<=l and j+k<=x;k++)
          nx[i-1][j+k]+=dp[i][j]*M(i-1)*M(l-k-1);
      }
    }

    for(int i=0;i<MAX;i++)
      for(int j=0;j<MAX;j++)
        dp[i][j]=nx[i][j],nx[i][j]=0;
  }


  M ans{0};
  for(int i=1;i<=n;i++)
    ans+=dp[i][x];
  cout<<ans<<endl;
  return 0;
}