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 int long long
signed main() {
//        ios :: sync_with_stdio(false);
//        cin.tie(0); cout.tie(0);
        int t; cin >> t;
        while(t--) {
                int n, m; cin >> n >> m;
                int arr[n], brr[n];
                for (int i = 0; i < n; i++) {
                        cin >> arr[i];
                }
                for (int i = 0; i < n; i++) {
                        cin >> brr[i];
                }
                int mx = 1e18, answer = 1e18;
                int one[n] = {}, two[n] = {};
                int cnt = 0, a = 0, b = 0;
                for (int i = 0; i < n; i++) {
                        if (arr[i] < m && brr[i] == 1) {
                                one[a] = arr[i];
                                cnt += arr[i];
                                a += 1;
                        } else if (arr[i] < m && brr[i] == 2) {
                                two[b] = arr[i];
                                cnt += arr[i];
                                b += 1;
                        } else {
                                mx = min(mx, brr[i]);
                        }
                }
                if (cnt < m || mx < 3) {
                        if (mx < 3) printf("%lld\n", mx);
                        else printf("-1\n");
                        continue;
                }
                auto compare = [&] (int a, int b) {
                        return a > b;
                };
                if (a) sort(one, one + a, compare);
                if (b) sort(two, two + b, compare);
                for (int i = 1; i < a; i++) one[i] += one[i - 1];
                for (int i = 1; i < b; i++) two[i] += two[i - 1];
                auto binarysearch = [&] (int need) {
                        int l = 0, r = a - 1;
                        int ans = -1;
                        while(l <= r) {
                                int mid = (l + r) / 2;
                                if (one[mid] >= need) {
                                        ans = mid;
                                        r = mid - 1;
                                } else {
                                        l = mid + 1;
                                }
                        }
                        return ans;
                };
                int p = binarysearch(m);
                if (p >= 0) answer = min(answer, p + 1);
                for (int i = 0; i < b; i++) {
                        int need = m - two[i];
                        int d = binarysearch(need);
                        if (d >= 0) {
                                d += (1 + (i + 1) * 2);
                                answer = min(answer, d);
                        }
                }
                cout << answer << endl;
        }
        return 0;
}