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.
import java.io.*;
import java.util.*;

final public class D {

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int tst = Integer.parseInt(br.readLine());
        while(tst-->0){
            String[] str = br.readLine().split(" ");
            int n = Integer.parseInt(str[0]), m = Integer.parseInt(str[1]);
            StringTokenizer st1 = new StringTokenizer(br.readLine());
            StringTokenizer st2 = new StringTokenizer(br.readLine());
            ArrayList<Integer> lst1 = new ArrayList<>();
            ArrayList<Integer> lst2 = new ArrayList<>();
            int b = 0, a = 0;
            for(int i = 0; i<n; i++){
                a = Integer.parseInt(st1.nextToken());
                b = Integer.parseInt(st2.nextToken());
                if(b == 1) lst1.add(a);
                else lst2.add(a);
            }
            lst1.sort(Collections.reverseOrder());
            lst2.sort(Collections.reverseOrder());
            int l1 = lst1.size(), l2 = lst2.size();
            long[] pref1 = new long[l1+1];
            long[] pref2 = new long[l2+1];
            for(int i = 1; i<=l1; i++){
                pref1[i] = pref1[i-1] + lst1.get(i-1);
            }
            for(int i = 1; i<=l2; i++){
                pref2[i] = pref2[i-1] + lst2.get(i-1);
            }
            int ans = Integer.MAX_VALUE, p2 = 0;
            while(p2+1<=l2 && pref2[p2]<m) p2++;
            if(pref2[p2]>=m)
            ans = 2*p2;
            for(int i = 1; i<=l1; i++){
                int l = -1, r = l2+1;
                while(l+1<r){
                    int mid = (l+r)/2;
                    if(pref1[i]+pref2[mid]>=m) r = mid;
                    else l = mid;
                }
                if(r!=l2+1){
                    ans = Math.min(ans, i + 2*r);
                }
            }
            if(ans == Integer.MAX_VALUE) sb.append(-1).append('\n');
            else sb.append(ans).append('\n');
        }
        System.out.println(sb);
    }
}
