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.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.StringTokenizer;
import java.util.stream.Stream;

public class D{
	
	static class InputOutput{
		static BufferedReader br;
		static StringTokenizer token;
		
		InputOutput(){
			br=new BufferedReader(new InputStreamReader(System.in));
			token=new StringTokenizer("");
		}
		
		static public int getInt() throws IOException{
			if(!token.hasMoreTokens())
				token=new StringTokenizer(br.readLine());
			return Integer.parseInt(token.nextToken());
		}
		static public int[] getIntArray(int n) throws IOException{
			token=new StringTokenizer(br.readLine());
			int arr[]=new int[n];
			for(int i=0;i<n;i++)
				arr[i]=Integer.parseInt(token.nextToken());
			return arr;
//			return Stream.of(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
		}
		static public int[] getIntArray(int size,int value) throws IOException{
			int arr[]=new int[size];
			Arrays.fill(arr, value);
			return arr;
		}

		static public int hcf(int a,int b) {
			return (a%b==0) ? b : hcf(b,a%b);
		}
		static public long lcm(int a,int b) {
			return ((long)a*b)/hcf(a,b);
		}
	}

	public static void main(String ar[]) throws IOException{

		InputOutput io=new InputOutput();
		int t=io.getInt();
		while(t--!=0) {
			int n=io.getInt();
			int m=io.getInt();
			int a[]=io.getIntArray(n);
			int b[]=io.getIntArray(n);
			ArrayList<Integer> first=new ArrayList();
			ArrayList<Integer> second=new ArrayList();
			for(int i=0;i<n;i++)
				if(b[i]==1)
					first.add(a[i]);
				else
					second.add(a[i]);
			
			Comparator<Integer> comp=(Integer a1,Integer a2)->a2-a1;
			Collections.sort(first,comp);
			Collections.sort(second,comp);
			int i=0,j=0;
			while((i<first.size() || j<second.size()) && m>0) {
				if(i==first.size())
					m=m-second.get(j++);
				else if(j==second.size())
					m=m-first.get(i++);
				else if(first.get(i)>=m)
					m=m-first.get(i++);
				else if(first.get(i)>=second.get(j))
					m=m-first.get(i++);
				else if(i+1<first.size() && first.get(i)+first.get(i+1)>=second.get(j))
					m=m-first.get(i++);
				else 
					m=m-second.get(j++);
			}
			
			if(m<=0)
				System.out.println((long)i*1+(long)j*2);
			else
				System.out.println(-1);
		}
	}
}