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.util.*;

import java.io.*;
public class Main 
{
    public static void main(String[] args) throws IOException
    {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        
        int t = in.nextInt();
        int go = 1;
        while(t-- != 0)
        {
        	int n = in.nextInt();
        	int m = in.nextInt();
        	Node[] nodes = new Node[n];
        	for(int i = 0; i < n; i++)
        		nodes[i] = new Node();
        	for(int i = 0; i < n; i++)
        		nodes[i].n1 = in.nextInt();
        	for(int i = 0; i < n; i++)
        		nodes[i].n2 = in.nextInt();
        	for(int i = 0; i < n; i++)
        		nodes[i].n3 = (float)nodes[i].n1 / nodes[i].n2;
        	Arrays.sort(nodes, new Comparator<Node>() {
				@Override
				public int compare(Node o1, Node o2) {
					// TODO 自动生成的方法存根
					if(o1.n3 == o2.n3)
						return 0;
					else if(o1.n3 > o2.n3)
						return -1;
					else 
						return 1;
				}
			});
//        	for(int i = 0; i < nodes.length; i++)
//        		System.out.println(nodes[i].toString());
        	int n1 = 0;
        	int n2 = 0;
        	for(int i = 0; i < n; i++)
        	{
        		n1 += nodes[i].n1;
        		n2 += nodes[i].n2;
        		//System.out.println("--->" + n1 + " " + n2);
        		if(n1 >= m)
        		{
        			if(nodes[i].n2 == 2)
        			{
        				int p1 = i - 1;
        				for(;p1 >= 0 && nodes[p1].n2 == 2; p1--);
        				if(p1 >= 0 && n1 - nodes[p1].n1 >= m)
        				{
        					//System.out.println("get到了");
        					n1 -= nodes[p1].n1;
        					n2--;
        				}
        				else
        				{
        					//System.out.println("没get到");
        					int p2 = i + 1;
            				for(;p2 < nodes.length && nodes[p2].n2 == 2; p2++);
            				if(p2 < nodes.length && n1 - nodes[i].n1 + nodes[p2].n1 >= m)
            				{
            					n1 = (int)(n1 - nodes[i].n1 + nodes[p2].n1);
            					n2--;
            				}
        				}
        				
        			}
        			break;
        		}
        	}
        	if(n1 < m)
        		out.println(-1);
        	else 
        		out.println(n2);
//        	if(go == 8 && n2 == 7)
//        	{
//        		out.println(n + " " + m);
//        		for(int i = 0; i < n; i++)
//        			out.print(nums[i][0] + " ");
//        		out.println();
//        		for(int i = 0; i < n; i++)
//        			out.print(nums[i][1] + " ");
//        		out.println();
//        	}
//        	else
//        		go++;
        		
        }
        
        
        out.flush();
    	
    }
    
    
}
class Node
{
	int n1;
	int n2;
	float n3;
	@Override
	public String toString()
	{
		return "" + n1 + " " + n2 + " " + n3;
	}
}

