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;

typedef long long ll;

int main()
{
    int T;
    cin >> T;

    while(T--)
    {
        int n, m;
        scanf("%d%d", &n, &m);

        vector <int> a(n);
        for(int& x : a) scanf("%d", &x);

        vector <vector<int>> A(2);
        for(int i = 0; i < n; i++)
        {
            int b;
            scanf("%d", &b);
            A[b - 1].push_back(a[i]);
        }

        for(int b = 0; b < 2; b++)
            sort(A[b].rbegin(), A[b].rend());

        vector <vector <ll>> prefix(2);
        for(int b = 0; b < 2; b++)
            prefix[b].assign(A[b].size() + 1, 0);

        for(int b = 0; b < 2; b++)
            for(int i = 0; i < A[b].size(); i++)
                prefix[b][i + 1] = prefix[b][i] + A[b][i];

        if(prefix[0][A[0].size()] + prefix[1][A[1].size()] < m)
        {
            puts("-1");
            continue;
        }

        int answer = A[0].size() + 2 * A[1].size();
        for(int cnt0 = A[0].size(), cnt1 = 0; cnt0 >= 0; cnt0--)
        {
            while(cnt1 <= A[1].size() && prefix[0][cnt0] + prefix[1][cnt1] < m)
                cnt1++;

            if(cnt1 <= A[1].size())
                answer = min(answer, cnt0 + 2 * cnt1);
        }

        printf("%d\n", answer);
    }
}
