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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;
 
#define Inf 1e9
 
vector<int> aux;
vector<long long> a[3];
 
int cb(int x) {
    int st = 0, dr = a[2].size() - 1, m;
    while (st <= dr) {
        m = (st + dr) / 2;
        if (a[2][m] >= x)
            dr = m - 1;
        else st = m + 1;
    }
    return st;
}
 
void solve() {
    int n, m, i, rez = Inf, x, ind1, ind2;
    a[1].resize(0);
    a[2].resize(0);
    aux.resize(0);
    cin >> n >> m;
    for (i = 1; i <= n; i++) {
        cin >> x;
        aux.push_back(x);
    }
    for (i = 1; i <= n; i++) {
        cin >> x;
        a[x].push_back(aux[i - 1]);
    }
    sort(a[1].rbegin(), a[1].rend());
    sort(a[2].rbegin(), a[2].rend());
 
    for (i = 1; i < a[1].size(); i++) 
        a[1][i] += a[1][i - 1];
    for (i = 1; i < a[2].size(); i++)
        a[2][i] += a[2][i - 1];
 
    int memory = m;
    ind2 = cb(memory);
    if (ind2 != a[2].size())
        rez = min(rez, 2 * (ind2 + 1));
    for (ind1 = 0; ind1 < a[1].size(); ind1++) {
        memory = m - a[1][ind1];
        if (memory > 0) {
            ind2 = cb(memory);
            if (ind2 != a[2].size())
                rez = min(rez, 1 * (ind1 + 1) + 2 * (ind2 + 1));
        }
        else {
            rez = min(rez, 1 * (ind1 + 1));
            break;
        }
    }
    if (rez == Inf)
        cout << -1;
    else
        cout << rez;
    cout << '\n';
}
 
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--)
        solve();
    return 0;
}