#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc = 1e9+7;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {return pow(modc-2);}
    mint& operator/=(const mint& a){ return (*this) *= a.inv();}
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{ return x == a.x;}
    bool operator != (const mint& a) const{ return x != a.x;}
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint &m) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){ return x;}
};

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    ll N, M, K, A;
    cin >> N >> K;
    M = N*100;
    //dp(i, j, k) = i番目まででj人選んで総和がkであるもの
    vector dp(N+1, vector(N+1, vector<mint>(M+1)));
    
    dp[0][0][0] = 1;
    for (int i=1; i<=N; i++){
        cin >> A;
        for (int j=0; j<=N; j++){
            for (int k=0; k<=M; k++){
                dp[i][j][k] += dp[i-1][j][k];
                if (j>=1 && k-A>=0) dp[i][j][k] += dp[i-1][j-1][k-A];
            }
        }
    }

    mint ans=0;
    //S/X>=K
    //S>=K*X
    for (int X=1; X<=N; X++){
        for (int S=0; S<=M; S++){
            if (S>=X*K) ans += dp[N][X][S];
        }
    }

    cout << ans << endl;

    return 0;
}