#include <bits/stdc++.h>
using namespace std;

const int N = 110, M = 25010;

int n, m, T;
int f[N][M];
int a[N];

int main()
{
    cin >> T;

    while(T -- )
    {
        int res = 0;

        memset(f,0,sizeof f);

        f[0][0] = 1;

        cin >> n;

        for(int i = 1; i <= n; i++) cin >> a[i];

        sort(a+1,a+1+n);

        m = a[n];

        for(int i = 1; i <= n; i++)
        {
            for(int j = 0; j <= m; j++)
            {
                f[i][j] = f[i-1][j];

                if(j >= a[i])
                f[i][j] += f[i][j-a[i]];
            }

        }

        //因为a数组已经排好序
        //如果a[i]不能用前i-1的面值凑出来，就需要保留
        for(int i = 1; i <= n; i++)
        {
            if(!f[i-1][a[i]]) res ++;
        }
        cout << res << endl;
    }
    return 0;
}