Polycarp often uses his smartphone. He has already installed n applications on it. Application with number i takes up a_i units of memory.

Polycarp wants to free at least m units of memory (by removing some applications).

Of course, some applications are more important to Polycarp than others. He came up with the following scoring system — he assigned an integer b_i to each application: 

  * b_i = 1 — regular application; 
  * b_i = 2 — important application. 



According to this rating system, his phone has b_1 + b_2 + … + b_n convenience points.

Polycarp believes that if he removes applications with numbers i_1, i_2, …, i_k, then he will free a_{i_1} + a_{i_2} + … + a_{i_k} units of memory and lose b_{i_1} + b_{i_2} + … + b_{i_k} convenience points.

For example, if n=5, m=7, a=[5, 3, 2, 1, 4], b=[2, 1, 1, 2, 1], then Polycarp can uninstall the following application sets (not all options are listed below): 

  * applications with numbers 1, 4 and 5. In this case, it will free a_1+a_4+a_5=10 units of memory and lose b_1+b_4+b_5=5 convenience points; 
  * applications with numbers 1 and 3. In this case, it will free a_1+a_3=7 units of memory and lose b_1+b_3=3 convenience points. 
  * applications with numbers 2 and 5. In this case, it will free a_2+a_5=7 memory units and lose b_2+b_5=2 convenience points. 



Help Polycarp, choose a set of applications, such that if removing them will free at least m units of memory and lose the minimum number of convenience points, or indicate that such a set does not exist.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m ≤ 10^9) — the number of applications on Polycarp's phone and the number of memory units to be freed.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the number of memory units used by applications.

The third line of each test case contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 2) — the convenience points of each application.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output on a separate line: 

  * -1, if there is no set of applications, removing which will free at least m units of memory; 
  * the minimum number of convenience points that Polycarp will lose if such a set exists. 

Example

Input


5
5 7
5 3 2 1 4
2 1 1 2 1
1 3
2
1
5 10
2 3 2 3 2
1 2 1 2 1
4 10
5 1 3 4
1 2 1 2
4 5
3 2 1 2
2 1 2 1


Output


2
-1
6
4
3

Note

In the first test case, it is optimal to remove applications with numbers 2 and 5, freeing 7 units of memory. b_2+b_5=2.

In the second test case, by removing the only application, Polycarp will be able to clear only 2 of memory units out of the 3 needed.

In the third test case, it is optimal to remove applications with numbers 1, 2, 3 and 4, freeing 10 units of memory. b_1+b_2+b_3+b_4=6.

In the fourth test case, it is optimal to remove applications with numbers 1, 3 and 4, freeing 12 units of memory. b_1+b_3+b_4=4.

In the fifth test case, it is optimal to remove applications with numbers 1 and 2, freeing 5 units of memory. b_1+b_2=3.
    #include <bits/stdc++.h>
    using namespace std;
     
    const int NMAX = 200005;
    const long long INF = 1e12;
    int t, n;
    long long a[NMAX], m, sum[NMAX];
     
     
    int main() {
        cin >> t;
        while(t--) {
            cin >> n >> m;
            for(int i = 1; i <= n; i++) {
                cin >> a[i];
            }
            long long sol = INF, s = 0;
            vector<long long> v1;
            vector<long long> v2;
            for(int i = 1; i <= n; i++) {
                long long x;
                cin >> x;
                if(x == 1) {
                    v1.push_back(a[i]);
                } else {
                    v2.push_back(a[i]);
                }
            }
            sort(v1.begin(), v1.end());
            sort(v2.begin(), v2.end());
            reverse(v1.begin(), v1.end());
            reverse(v2.begin(), v2.end());
            if(v2.size() > 0) {
                sum[0] = v2[0];
            }
            for(int i = 1; i < (int)v2.size(); i++) {
                sum[i] = sum[i - 1] + v2[i];
            }
            for(int i = -1; i < (int)v1.size(); i++) {
                if(i >= 0) {
                    s += v1[i];
                }
                int st = 0;
                int dr = v2.size() - 1;
                if(s >= m) {
                    sol = min(sol, 1LL * i + 1);
                }
                while(st <= dr) {
                    int mid = st + (dr - st) / 2;
                    if(s + sum[mid] >= m) {
                        sol = min(sol, 1LL * i + 1 + 2 * (1LL * mid + 1));
                        dr = mid - 1;
                    } else {
                        st = mid + 1;
                    }
                }
            }
            if(sol == INF) {
                cout << -1 << '\n';
            } else {
                cout << sol << '\n';
            }
        }
        return 0;
    }