// Original C++ solution (slower):
#include <bits/stdc++.h>

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)

#define rep(i,n) for (int i = (0); i < (n); ++i)

#define rng(x) (x).begin(),(x).end()

#define rrng(x) (x).rbegin(),(x).rend()

#define show(x) cout << #x << " = " << (x) << endl;

#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl;

#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl;

#define fi first

#define se second

#define pb push_back

#define eb emplace_back

#define sz(x) (int)(x).size()

#define bn(x) ((1<<x)-1)

#define dup(x,y) (((x)+(y)-1)/(y))

#define uni(x) x.erase(unique(rng(x)),x.end())

#define SP << " " <<

using namespace std;

using ll = long long;

using vi = vector<int>;

using vvi = vector<vi>;

using vll = vector<ll>;

using pii = pair<int, int>;

using tiii = tuple<int, int, int>;

const int mod = 1000000007;

const double EPS = 1e-9;

const int INF = 1<<30;

const ll INFLL = 1LL<<60;

//cout << fixed << setprecision(10);



template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }



// X = max({x[0],...,x[n-1],a})

// y[i] = x[i] - a

//ll dp[i][j]: y[0],..,y[i-1] から0枚以上選んで合計をj-n*Xにするような選び方の総数



ll dp[55][100000];

ll x[55];



int main () {

  int n, a;

  cin >> n >> a;

  rep(i,n) cin >> x[i];

  int X = *max_element(x,x+n);



  vi y(n);

  rep(i,n) y[i] = x[i] - a;



  rep(i,55)rep(j,6050) dp[i][j] = 0;

  dp[0][n*X] = 1;

  rep(i,50) {

    rep(j,5050) {

      if (dp[i][j] == 0) continue;

      dp[i+1][j] += dp[i][j];

      dp[i+1][j+y[i]] += dp[i][j];

    }

  }



  ll ans = dp[n][n*X] - 1;

  cout << ans << endl; 

  return 0;

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)

#define rep(i,n) for (int i = (0); i < (n); ++i)

#define rng(x) (x).begin(),(x).end()

#define rrng(x) (x).rbegin(),(x).rend()

#define show(x) cout << #x << " = " << (x) << endl;

#define show2(x,y) cout << #x << " = " << (x) << ", " << #y << " = " << (y) << endl;

#define show3(x,y,z) cout << #x << " = " << (x) << ", " << #y << " = " << (y) <<  ", " << #z << " = " << (z) << endl;

#define fi first

#define se second

#define pb push_back

#define eb emplace_back

#define sz(x) (int)(x).size()

#define bn(x) ((1<<x)-1)

#define dup(x,y) (((x)+(y)-1)/(y))

#define uni(x) x.erase(unique(rng(x)),x.end())

#define SP << " " <<

using namespace std;

using ll = long long;

using vi = vector<int>;

using vvi = vector<vi>;

using vll = vector<ll>;

using pii = pair<int, int>;

using tiii = tuple<int, int, int>;

const int mod = 1000000007;

const double EPS = 1e-9;

const int INF = 1<<30;

const ll INFLL = 1LL<<60;

//cout << fixed << setprecision(10);



template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }



ll x[55];

ll y[55];

ll dp[55][5050];



int main () {

  ll n, a;

  cin >> n >> a;

  rep(i,n) cin >> x[i];

  ll X = *max_element(x,x+n);

  rep(i,n) y[i] = x[i] - a;

  

  dp[0][n*X] = 1;

  rep(i,n+1) {

    rep(j,2*n*X+1) {

      if (dp[i][j] == 0) continue;

      dp[i+1][j] += dp[i][j];

      dp[i+1][j+y[i]] += dp[i][j];

    }

  }



  ll ans = dp[n][n*X] - 1;

  cout << ans << endl;

  return 0;

}

// Fastest known solution for this problem: