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.
// Working program with FastReader 
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class D_Cleaning_the_Phone{
    public static void main(String[] args) {
       FastReader sc=new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int test = sc.nextInt();
        for(int t=0;t<test;t++){
            int n = sc.nextInt();
            int k = sc.nextInt();
            long ans = 0;
            int[] array = new int[n];
            int[] value = new int[n];
            for(int i =0;i<n;i++){
                array[i] = sc.nextInt();
                ans += array[i];
            }
            for(int i=0;i<n;i++){
                value[i] = sc.nextInt();
            }
            if(ans<k){
                out.println(-1);
            }
            else{
                int result = solve(array,value,n,k);
                out.println(result);
            }
        }
        out.close();
    }

    public static int solve(int[] array,int[] value,int n,int k){
        List<Integer> one = new ArrayList<>();
        List<Integer> two = new ArrayList<>();
        for(int i =0;i<n;i++){
            if(value[i]==1){
                one.add(array[i]);
            }
            else{
                two.add(array[i]);
            }
        }
        Collections.sort(one,Collections.reverseOrder());
        Collections.sort(two,Collections.reverseOrder());
        int[] prefix1 = new int[one.size()+1];
        int[] prefix2 = new int[two.size()+1];
        for(int i=0;i<one.size();i++){
            prefix1[i+1] = prefix1[i]+one.get(i);
        }
        for(int i=0;i<two.size();i++){
            prefix2[i+1] = prefix2[i]+two.get(i);
        }
        int ans = prefix1.length + 2*prefix2.length;
        for(int i =0;i<prefix1.length;i++){
            int remaining = k-prefix1[i];
            int tempans = i;
            int second = binarySearch(prefix2, remaining);
            tempans += 2*second;
            ans = Math.min(ans,tempans);
            // if(second==prefix2.length){
            //     continue;
            // }
            // else{
                
            // }
        }
        return ans;
    }

    public static int binarySearch(int[] prefix2,int target){
        int left = 0;
        int right = prefix2.length-1;
        while(left<=right){
            int mid = left+(right-left)/2;
            if(prefix2[mid]==target){
                return mid;
            }
            else if(prefix2[mid]<target){
                left = mid+1;
            }
            else{
                right = mid-1;
            }
        }
        if(left<prefix2.length && prefix2[left]<target){
            return left+1;
        }
        else return left;
    }

    // public static int solve(int[] array,int[] value,int n,int k){
    //     List<Integer> one = new ArrayList<>();
    //     List<Integer> two = new ArrayList<>();
    //     for(int i =0;i<n;i++){
    //         if(value[i]==1){
    //             one.add(array[i]);
    //         }
    //         else{
    //             two.add(array[i]);
    //         }
    //     }
    //     Collections.sort(one,Collections.reverseOrder());
    //     Collections.sort(two,Collections.reverseOrder());
    //     int ans = 0;
    //     int i=0;
    //     int j=0;
    //     int no=0;
    //     while(ans<k){
    //         if(i==one.size()){
    //             ans += two.get(j);
    //             no+=2;
    //             j++;
    //         }
    //         else if(j==two.size()){
    //             ans+=one.get(i);
    //             no++;
    //             i++;
    //         }
    //         else if(i==one.size()-1){
    //             if(one.get(i)>=two.get(j)){
    //                 ans+=one.get(i);
    //                 no+=1;
    //                 i++;
    //             }
    //             else{
    //                 if(ans+one.get(i)>=k){
    //                     ans+=one.get(i);
    //                     i++;
    //                     no+=1;
    //                 }
    //                 else{
    //                     ans+=two.get(j);
    //                     j++;
    //                     no+=2;
    //                 }
                    
    //             }
    //         }
    //         else{
    //             int temp1 = one.get(i);
    //             int temp2 = temp1+one.get(i+1);
    //             int temp3 = two.get(j);
    //             if(ans+temp1>=k){
    //                 ans+= temp1;
    //                 no++;
    //                 i++;
    //             }
    //             else if(ans+temp2>=k){
    //                 ans+=temp2;
    //                 no+=2;
    //                 i+=2;
    //             }
    //             else if(ans+temp3>=k){
    //                 ans+= temp3;
    //                 no+=2;
    //                 j++;
    //             }
    //             else if(ans+temp1+temp3>=k){
    //                 ans+=temp1+temp3;
    //                 no++;
    //                 no+=2;
    //                 i++;
    //                 j++;
    //             }
    //             else if(temp2>temp3){
    //                 ans+=temp2;
    //                 no+=2;
    //                 i+=2;
    //             }
    //             else{
    //                 ans+=temp3;
    //                 no+=2;
    //                 j++;
    //             }
    //         }
            

    //     }
    //     return no;
    // }
    
    public static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 

        public FastReader() 
        { 
            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; 
        } 
}
}
