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.*;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Main{
    private static int n, m;
    private static List<Integer> one, two;
    public static void main(String [] args) throws IOException{
        try(
                BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
                BufferedReader br = new BufferedReader(new InputStreamReader(System.in))
                ){
            int tc = Integer.parseInt(br.readLine());
            while(tc-- != 0){
                StringTokenizer st = new StringTokenizer(br.readLine());
                n = Integer.parseInt(st.nextToken());
                m = Integer.parseInt(st.nextToken());

                int [] a = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
                int [] b = Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
                int sumSize = Arrays.stream(a).sum();
                int sumPoint = Arrays.stream(b).sum();

                if(sumSize < m){
                    bw.write("-1\n");
                    continue;
                }else if(sumSize == m){
                    bw.write(sumPoint + "\n");
                    continue;
                }

                one = new ArrayList<>();
                two = new ArrayList<>();
                IntStream.range(0, n)
                        .forEach(i -> {
                            if(b[i] == 1) one.add(a[i]);
                            else two.add(a[i]);
                        });

                one.sort(Integer::compare);
                two.sort(Integer::compare);

                int idx1 = one.size() - 1, idx2 = two.size() - 1, remove = 0, lose = 0;
                while(idx1 > 0 && idx2 >= 0){
                    if(one.get(idx1) + remove >= m){
                        remove += one.get(idx1);
                        lose++;
                        break;
                    }
                    int sumOne = one.get(idx1) + one.get(idx1 - 1);
                    int sumTwo = two.get(idx2);
                    if(sumOne + remove < m && sumTwo + remove < m){
                        if(one.get(idx1) + two.get(idx2) + remove >= m){
                            lose += 3;
                            remove += one.get(idx1) + two.get(idx2);
                            break;
                        }
                    }
                    if(sumOne >= sumTwo){
                        remove += sumOne;
                        idx1 -= 2;
                    }else{
                        remove += sumTwo;
                        idx2 --;
                    }
                    lose += 2;
                    if(remove >= m) break;
                }
                while(remove < m && idx1 >= 0 && idx2 >= 0){
                    if(remove + one.get(idx1) >= m){
                        remove += one.get(idx1);
                        lose++;
                        continue;
                    }
                    if (one.get(idx1) >= two.get(idx2)){
                        remove += one.get(idx1);
                        idx1--;
                        lose++;
                    }else{
                        remove += two.get(idx2);
                        idx2--;
                        lose += 2;
                    }
                }
                while(remove < m && idx1 >= 0){
                    remove += one.get(idx1);
                    idx1--;
                    lose++;
                }
                while(remove < m && idx2 >= 0){
                    remove += two.get(idx2);
                    idx2--;
                    lose += 2;
                }
                bw.write(lose + "\n");

            }
            bw.flush();
        }
    }
}