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;
#define ll long long
#define int ll
#define oo (int)(1e17)

int32_t main() {
    cin.sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
    freopen("input1.txt", "r", stdin);
    freopen("output1.txt", "w", stdout);
#endif
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        int a[n + 1];
        int b[n + 1];
        for (int i = 1; i <= n; i++) {
            cin >> a[i];
        }
        vector<int> ones;
        vector<int> twos;
        for (int i = 1; i <= n; i++) {
            cin >> b[i];
            if (b[i] == 1) {
                ones.push_back(a[i]);
            }
            else {
                twos.push_back(a[i]);
            }
        }
        sort(ones.begin(), ones.end());
        sort(twos.begin(), twos.end());
        int pre1[n + 1];
        int pre2[n + 1];
        pre1[0] = 0;
        pre2[0] = 0;
        int aa = (int)(ones.size());
        int bb = (int)(twos.size());
        for (int i = 1; i <= aa; i++) {
            pre1[i] = pre1[i - 1] + ones.back();
            // cout << pre1[i] << " ";
            ones.pop_back();
        }
        for (int i = 1; i <= bb; i++) {
            pre2[i] = pre2[i - 1] + twos.back();
            // cout << pre2[i] << " ";
            twos.pop_back();
        }
        int mn = oo;
        for (int i = 0; i <= aa; i++) {
            auto it = lower_bound(pre2, pre2 + bb + 1, m - pre1[i]);
            if (it != pre2 + bb + 1) {
                mn = min(mn, (ll)(i + 2 * ((int)(it - pre2))));
            }
        }
        if (mn == oo) {
            cout << "-1" << endl;
        }
        else {
            cout << mn << endl;
        }
    }
}