样例都过了，下面只过了前五个点。思路是一旦判断到能进行元素反应，那么一定yes。因为翻倍后一定仍然满足整除，所以可以无限翻倍。 如果没有能元素反应的，判断和是否大于k。如果大于输出yes反之no。

#include <iostream>
#include <vector>

using namespace std;
#define int long long 
signed main() {
    int T;
    cin >> T;

    while (T--) {
        int n, k,fg=0;
        cin >> n >> k;

        vector<int> a(n);
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }
        for (int i = 0; i < n; ++i) {
            for (int j = i+1; j < n; ++j) {
                int product = a[i] * a[j];
                if (product % 147 == 0 || product % 154 == 0) {
                    fg=1;
                    goto qwe;
                    
                 }
                
        
                  
                
            }
        }
        qwe:
        if(fg==1) cout << "Yes" << endl;
        else{
        long long total = 0;
        for (int i = 0; i < n; ++i) {
            total += a[i];
        }

        if (total >= k) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
        }
    }

    return 0;
};