// Original C++ solution (slower):
#define SORT(v) sort(v.begin(), v.end())

#include <iostream>

#include <map>

#include <vector>

#include <queue>

#include <stack>

#include <cstdlib>

#include <algorithm>

#include <cstdarg>

#include <cstdio>

#include <cmath>

#include <numeric>

 

// #include "ane.cpp"

 

#define INF (int)1e9

#define INFLL (long long)1e18

#define NMAX 52

#define MMAX 200

#define MOD 100000

using namespace std;

 

// コメントアウトするとdb_printf()を実行しない

// #define DEBUG

 

 

 

 

 

//

//  ライブラリ

//

 

// frequent used aliases

typedef pair<int, int> p;

typedef long long ll;

 

// frequent used constants

static const int di[] = {-1, 0, 1, -1, 1, -1, 0, 1};

static const int dj[] = {-1, -1, -1, 0, 0, 1, 1, 1};

 

// デバッグ用printf

void db_printf(const char* format, ...){

    #ifndef DEBUG

    return;

    #endif

    va_list arg;

    va_start(arg, format);

    vprintf(format, arg);  // コンソールに出力

    va_end(arg);

}

 

// n次元配列の初期化。第２引数の型のサイズごとに初期化していく。

template<typename A, size_t N, typename T>

void Fill(A (&array)[N], const T &val){

    std::fill( (T*)array, (T*)(array+N), val );

}

 

// Union-Find Tree

class UnionFindTree{

    struct node{

        int par;

    };

    std::vector<node> T;

 

    public:

    void init(int num){

        T.resize(num+1);    // ignore T[0]

        for (int i = 1; i <= num; ++i)

        {

            T[i].par = i;

        }

    }

    void unite(int x, int y){

        T[find(y)].par = find(x);

    }

    int find(int x){

        if (T[x].par == x) return x;

        else return T[x].par = find(T[x].par);

    }

    bool same(int x, int y){

        return find(x) == find(y);

    }

};

 

// Segment Tree for Range Minimum Query

    // **********************************************************

    // *** important: all functions' variable,                ***

    // *** such as "index", "l", "r", etc., must be 0-origin. ***

    // **********************************************************

    // ********************************************

    // *** important: NMAX must be power of 2. ***

    // ********************************************

template<typename T>

class SegmentTree{

    private:

    ll N;

    T INF_VAL;

    T dat[NMAX * 2]; // 0 origin, A[i] = dat[i + N]

 

    T _query(ll l, ll r, ll l_responsible, ll r_responsible, ll i_dat){

        db_printf("query(%lld, %lld, %lld, %lld, %lld) called\n",

            l, r, l_responsible, r_responsible, i_dat);

        if (r < l_responsible || r_responsible < l) return INF_VAL;

        else if(l <= l_responsible && r_responsible <= r) return dat[i_dat];

        else return min(_query(l ,r, l_responsible, (l_responsible + r_responsible) / 2, i_dat * 2),

                        _query(l, r, (l_responsible + r_responsible) / 2 + 1, r_responsible, i_dat * 2 + 1));

    }

 

    public:

    void init(ll _N, T _inf_val){

        N = 1; while(N < _N) N *= 2;

        INF_VAL = _inf_val;

        Fill(dat, _inf_val);

    }

    T get(ll index){

        return dat[index + N];

    }

    void set(int index, T val){

        int i_dat = index + N;

        dat[i_dat] = val;

        for (i_dat /= 2; i_dat > 0; i_dat /= 2)

        {

            dat[i_dat] = min(dat[i_dat * 2], dat[i_dat * 2 + 1]);

        }

    }

    T query(ll l, ll r){

        return _query(l, r, 0, N - 1, 1);

    }

    void dump(){

        cout << "*** SegTree dump begin ***\n";

        cout << "N = " << N << ", INF_VAL = " << INF_VAL << endl;

        for (int i = 1; i < N * 2; i *= 2)

        {

            for (int j = i; j < i * 2; ++j)

            {

                if(dat[j] == INF_VAL) cout << "INF ";

                else cout << dat[j] << " ";

            }

            cout << endl;

        }

        cout << "*** SegTree dump end ***\n";

    }

};

 

// Binary Indexed Tree for Range Sum Query

    // *******************************************

    // *** important: all functions' variable, ***

    // *** such as "i", must be 1-origin.      ***

    // *******************************************

template<typename T>

class BinaryIndexedTree{

    private:

    ll N;

    T dat[NMAX + 1]; // 1 origin, A[i] = sum(i) - sum(i-1)

 

    public:

    void init(ll _N){

        N = _N;

        Fill(dat, 0);

    }

    void add(int i, T val){

        while(i <= N) {

            dat[i] += val;

            i += i & -i;

        }

    }

    T sum(ll i){

        T ret = 0;

        while(i > 0) {

            ret += dat[i];

            i -= i & -i; // set last HIGH bit to LOW

        }

        return ret;

    }

    T sum(ll left, ll right){

        return sum(right) - sum(left - 1);

    }

    void dump(){

        cout << "*** BITree dump begin ***\n";

        cout << "N = " << N << endl;

        for (int i = 1; i <= N; i *= 2)

        {

            cout << dat[i] << " ";

        }

        cout << "*** BITree dump end ***\n";

    }

};

 

//

// ライブラリ終了

//

 

 

 

 

 

 

 

ll N,M,K,A[NMAX],B,C,D,E;

ll dp[NMAX][NMAX * 100] = {};

ll ans = 0;

 

void solve(){

    // 解答アルゴリズム

    int offset = NMAX * 50;

    dp[0][offset] = 1LL;

    for (int i = 1; i < N + 1; ++i)

    {

        for (int j = 0; j < NMAX * 100; ++j)

        {

            if (j - A[i] < 0 || j - A[i] >= NMAX * 100) dp[i][j] = dp[i-1][j];

            else dp[i][j] = dp[i-1][j] + dp[i-1][j - A[i]];

        }

    }

 

    ans = dp[N][offset] - 1;// cancel obvious solution

}

void debug(){

    // デバッグ用出力

    cout << "N = " << N << ", M = " << M << ", A[i]:\n";

    for(int i = 0; i < N; ++i){

        cout << A[i] << " ";

    }

    cout << endl;

}

void answer(){

    // 解答出力

    printf("%lld\n", ans);

}

int main(int argc, char const *argv[])

{

    // 入力の読み込み，番兵法

 

    // Fill(dp, -1);

    scanf("%lld%lld", &N,&M);

    for (int i = 1; i <= N; ++i)

    {

        scanf("%lld", &A[i]);

        A[i] -= M;

    }

    solve();

    #ifdef DEBUG

    debug();

    #endif

    answer();

 

    return 0;

}

// Optimized solution to the same problem (faster):
#include <cstdio>

 

using namespace std;

 

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

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

 

const int N_MAX = 50;

const int DP0 = 2500;

 

typedef long long ll;

 

int N, A;

int x[N_MAX];

 

ll dp[DP0 * 2 + 1];

 

void solve() {

  dp[DP0] = 1;

  REP(i, N) {

    x[i] -= A;

    if (x[i] < 0) {

      for (int j = -DP0; j <= DP0 + x[i]; j++) {

        dp[DP0 + j] += dp[DP0 + j - x[i]];

      }

    } else {

      for(int j = DP0; j >= -DP0 + x[i]; j--) {

        dp[DP0 + j] += dp[DP0 + j - x[i]];

      }

    }

  }

  printf("%lld\n", dp[DP0] - 1);

}

 

void input() {

  scanf("%d%d", &N, &A);

  REP(i, N) scanf("%d", x + i);

}

 

int main() {

  input();

  solve();

  return 0;

}

// Fastest known solution for this problem: