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;

#define int long long

#define rep(i,n) for(int i=0;i<(n);i++)
#define pb push_back
#define eb emplace_back
#define all(v) (v).begin(),(v).end()
#define fi first
#define se second

using vint=vector<int>;
using pint=pair<int,int>;
using vpint=vector<pint>;

template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

template<class A,class B>
ostream& operator<<(ostream& ost,const pair<A,B>&p){
    ost<<"{"<<p.first<<","<<p.second<<"}";
    return ost;
}

template<class T>
ostream& operator<<(ostream& ost,const vector<T>&v){
    ost<<"{";
    for(int i=0;i<v.size();i++){
        if(i)ost<<",";
        ost<<v[i];
    }
    ost<<"}";
    return ost;
}

inline int topbit(unsigned long long x){
    return x?63-__builtin_clzll(x):-1;
}

inline int popcount(unsigned long long x){
    return __builtin_popcountll(x);
}

inline int parity(unsigned long long x){
    return __builtin_parity(x);
}

template<uint32_t mod>
struct ModInt{
    uint32_t a;
    ModInt& s(uint32_t vv){
        a=vv<mod?vv:vv-mod;
        return *this;
    }

    ModInt(int64_t x=0){s(x%mod+mod);}

    ModInt& operator+=(const ModInt &x){return s(a+x.a);}
    ModInt& operator-=(const ModInt &x){return s(a+mod-x.a);}
    ModInt& operator*=(const ModInt &x){
        a=uint64_t(a)*x.a%mod;
        return *this;
    }
    ModInt& operator/=(const ModInt &x){
        *this*=x.inv();
        return *this;
    }

    ModInt operator+(const ModInt &x)const{return ModInt(*this)+=x;}
    ModInt operator-(const ModInt &x)const{return ModInt(*this)-=x;}
    ModInt operator*(const ModInt &x)const{return ModInt(*this)*=x;}
    ModInt operator/(const ModInt &x)const{return ModInt(*this)/=x;}
    bool operator==(const ModInt &x)const{return a==x.a;}
    bool operator!=(const ModInt &x)const{return a!=x.a;}
    bool operator<(const ModInt &x)const{return a<x.a;}

    ModInt operator-()const{return ModInt()-*this;}
    ModInt pow(int64_t n)const{
        ModInt res(1),x(*this);
        while(n){
            if(n&1)res*=x;
            x*=x;
            n>>=1;
        }
        return res;
    }

    ModInt inv()const{return pow(mod-2);}
};

template<uint32_t mod>
istream& operator>>(istream& in,ModInt<mod>& a){
    return (in>>a.a);
}
template<uint32_t mod>
ostream& operator<<(ostream& out,const ModInt<mod>& a){
    return (out<<a.a);
}
using mint=ModInt<1000000007>;

template<class Mint,int32_t lg>
struct ModIntTable{
    int N;
    vector<Mint>facts,finvs,invs;
    ModIntTable():N(1<<lg),facts(N),finvs(N),invs(N){
        const uint32_t mod=Mint(-1).a+1;
        invs[1]=1;
        for(int i=2;i<N;i++)invs[i]=invs[mod%i]*(mod-mod/i);

        facts[0]=1;
        finvs[0]=1;
        for(int i=1;i<N;i++){
            facts[i]=facts[i-1]*i;
            finvs[i]=finvs[i-1]*invs[i];
        }
    }
    inline Mint fact(int n)const{return facts[n];}
    inline Mint finv(int n)const{return finvs[n];}
    inline Mint inv(int n)const{return invs[n];}
    inline Mint binom(int n,int k)const{
        if(n<0||k<0||k>n)return 0;
        return facts[n]*finvs[k]*finvs[n-k];
    }
    inline Mint perm(int n,int k)const{
        if(n<0||k<0||k>n)return 0;
        return facts[n]*finvs[n-k];
    }
    inline Mint catalan(int n){
        return facts[2*n]*finvs[n+1]*finvs[n];
    }
};
ModIntTable<mint,20>mt;

int N,X;
int L[111];

mint dp[510][510];

mint po[555][555];
int cnt[555];
signed main(){
    for(int i=0;i<555;i++){
        po[i][0]=1;
        for(int j=1;j<555;j++)po[i][j]=po[i][j-1]*i;
    }
    cin>>N>>X;
    rep(i,N)cin>>L[i];
    rep(i,N)cnt[L[i]]++;


    dp[0][0]=1;
    for(int i=X;i>=0;i--){
        for(int j=X+1;j>=0;j--){
            for(int k=0;k<=X+1;k++){
                if(dp[j][k]==0)continue;
                for(int l=1;k+(i+1)*l<=X+1;l++){
                    mint tmp=dp[j][k]*mt.binom(j+l,j);
                    dp[j+l][k+(i+1)*l]+=tmp;
                }
            }
        }
        if(i){
            for(int j=0;j<=X+1;j++){
                for(int k=0;k<=X+1;k++){
                    int w=k-j*i;
                    if(w<0||w>X)continue;
                    dp[j][k]*=po[w][cnt[i]];
                }
            }
        }
    }



    mint ans=0;
    for(int j=1;j<=X+1;j++){
        if(j&1)ans+=dp[j][X+1];
        else ans-=dp[j][X+1];
    }
    cout<<ans<<endl;
    return 0;
}