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.
/*****************************
*  Author :: Πρασαννα Κ. Ροι  *
*****************************/

/***********************************
* Unus pro omnibus, omnes pro uno  *
***********************************/
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp> 
#include<ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 
using namespace std;
using namespace std::chrono;

#define __AcHiLlEs ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define __AnAkLuSmOs freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define int long long
#define span(a) begin(a), end(a)

template<typename t>
using vc = vector<t>;

void solve() {
    int n, m, sz1(0), sz2(0);
    cin >> n >> m;
    vc<array<int, 2>> v(n), a, b;
    for (auto &i: v) cin >> i[0];
    for (auto &i: v) cin >> i[1];
    for (auto &i: v) {
        if (i[1] == 2) a.push_back(i), sz1++;
        else b.push_back(i), sz2++;
    }
    sort(span(a), greater<array<int, 2>>());
    sort(span(b), greater<array<int, 2>>());
    int i = 0, j = 0, sum(0), res(0);
    while (sum < m and (i < sz1 or j < sz2)) {
        int sum1(0), sum2(0);
        if (j < sz2 and (sum + b[j][0]) >= m) return void(cout << res + 1 << "\n");
        if (i < sz1 and (sum + a[i][0]) >= m) return void(cout << res + 2 << "\n");
        if (i < sz1) sum1 = a[i][0];
        if (j < sz2) sum2 = b[j][0] * 2;
        if (sum1 > sum2) sum += sum1, i++, res++;
        else sum += (sum2 / 2), j++, res++;
    }
    if (sum < m) cout << -1 << "\n";
    else cout << res << "\n";
}

signed main() {
    // auto start = high_resolution_clock::now();
    #ifndef ONLINE_JUDGE 
        __AnAkLuSmOs    
    #endif

    __AcHiLlEs

    int t(1);
    cin >> t;
    for (int i = 1; i <= t; /*cout << "Case " << i << ": ",*/ solve(), i++);
    // auto stop = high_resolution_clock::now();
    // double duration = duration_cast<microseconds>(stop - start).count();
    // cout << fixed << setprecision(4) << duration / 1000000 << "\n";  
    return 0;
}