//#include <atcoder/all>
//using namespace atcoder;
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i, n) for(int i=0; i<n; i++)
#define REPR(i, n) for(int i=n-1; i>=0; i--)
#define FOR(i, m, n) for(int i=m; i<n; i++)
#define ALL(v) v.begin(), v.end()
#define SIZE(x) int(x.size()) 
#define bit(n) (1LL<<(n))
#define FIX(d) fixed << setprecision(d)
using P = pair<int, int>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = 2e9;
const ll LINF = (1LL<<60);

const int mod = 1000000007;
//const int mod = 998244353;
struct mint {
  ll x;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

mint dp[105][20005];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n,k; cin >> n >> k;
  vector<int> a(n);
  REP(i,n) cin >> a[i], a[i] -= k;
  
  dp[0][10000] = 1;
  REP(i,n) REP(j,20001){
    dp[i+1][j] += dp[i][j];
    if(j+a[i]<=20000) dp[i+1][j+a[i]] += dp[i][j];
  }

  mint ans = 0;
  dp[n][10000] -= 1;
  FOR(j,10000,20000) ans += dp[n][j];
  cout << ans << endl;
  return 0;
}