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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.StringTokenizer;


public class Main {
    static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }

    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        int t = sc.nextInt();

        while (t-- != 0){
            int n = sc.nextInt();
            int m = sc.nextInt();

            ArrayList <Integer> a1 = new ArrayList<>();
            ArrayList<Integer> a2 = new ArrayList<>();

            int a [] =new int[n];
            for(int i=0;i<n;i++){
                a[i] = sc.nextInt();
            }

            for(int i=0;i<n;i++){
                int x =sc.nextInt();
                if(x == 1){
                    a1.add(a[i]);
                }else{
                    a2.add(a[i]);
                }
            }

            Collections.sort(a1,Collections.reverseOrder());
            Collections.sort(a2,Collections.reverseOrder());

            int  i =0;
            int j =0;
            int ans =0;
            while(i < a1.size() || j<a2.size()){
                int x1 = -1000000001;
                int x2 = -1000000001;
                int y1 = -1000000001;
                if(i < a1.size()){
                    x1 = a1.get(i);
                }
                if(i +1 < a1.size()){
                    x2 =  a1.get(i+1);
                }
                if(j < a2.size()){
                    y1 = a2.get(j);
                }

                if(m - x1 <= 0){
                    m = 0;
                    ans += 1;
                    break;
                }
                if(m - y1 <= 0 || m - x1- x2 <= 0){
                    m =0;
                    ans +=2;
                    break;
                }
                if(m - y1 - x1 <= 0){
                    m=0;
                    ans += 3;
                    break;
                }
                if(y1 > x1+x2){
                    m -= y1;
                    ans +=2;
                    j ++;
                }else{
                    m-= (x1 + x2);
                    ans += 2;
                    i+=2;
                }

                if(m <= 0){
                    break;
                }

            }
            if(m <= 0)
                System.out.println(ans);
            else
                System.out.println(-1);

        }
    }
}
