//本题为计数类DP
//状态表示：f(i,j)表示从前0~i种选，货币总面值为j
/*
状态计算：以第i种货币选的数量为划分，分别划分为0、1、2、...、INF种
f(i,j) = f(i-1,j)          0
f(i,j) = f(i-1,j-w[i])     1
f(i,j) = f(i-1,j-2*w[i])   2
......
可发现：f(i,j-w[i]) = f(i-1,j-w[i]) + f(i-1,j-2*w[i]) + f(i-1,j-3*w[i]) + ...
故最终状态转移方程：f(i,j) = f(i-1,j) + f(i,j-w[i])
*/

#include <iostream>

using namespace std;

typedef long long LL;

const int N = 30 , M = 10010;

int n,m;
int w[N];
LL f[N][M];

int main(){

    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%d",&w[i]);
    }

    f[0][0] = 1;  //初始啥都不选即为一种选法
    //开始DP
    for(int i=1;i<=n;i++){
        for(int j=0;j<=m;j++){
            f[i][j] = f[i-1][j];
            if(j>=w[i]){
                f[i][j] += f[i][j-w[i]];
            }
        }
    }

    printf("%lld\n",f[n][m]);

    return 0;
}