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.math.*;
import java.security.*;
import java.text.*;
import java.time.LocalTime;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;


public class B {
	

	public static void main(String[] args) {
    	FastScanner scan = new FastScanner();
    	
    	int t = scan.nextInt();
    		for(int tt=0; tt<t; tt++) {
    			int n = scan.nextInt(), m = scan.nextInt();
    			int [] a = scan.readArray(n), b = scan.readArray(n);
    			ArrayList<Integer> good = new ArrayList<>(), bad = new ArrayList<>();
    			for(int i=0; i<n; i++) {
    				if(b[i] == 2) good.add(a[i]);
    				else bad.add(a[i]);
    			}
    			Collections.sort(good); Collections.sort(bad);
    			Collections.reverse(good); Collections.reverse(bad);
    			
    			int p1 = 0, p2 = 0;
    			int currMemory = m, currPoints = 0;
    			int ans = Integer.MAX_VALUE;
    			
    			while(p1 < good.size() && currMemory > 0) {
    				currMemory -= good.get(p1);
    				currPoints += 2; p1++;
    			}
    			p1 --;
    			if(currMemory <= 0) ans = Math.min(currPoints, ans);
    			
    			for(int i = 0; i<bad.size(); i++) {
    				currMemory -= bad.get(i);
    				currPoints += 1; 
    				
    				while(p1 >= 0 && currMemory + good.get(p1) <= 0) {
    					currMemory += good.get(p1);
    					p1 --; currPoints -= 2;
    				}
    				if(currMemory <=0) ans = Math.min(ans, currPoints);
    			}
    			if(ans == Integer.MAX_VALUE) ans = -1;
    			System.out.println(ans);
    		}
	}
	

	
	  public static void sort(int [] a) {
	    	ArrayList<Integer> b = new ArrayList<>();
	    	for(int i: a) b.add(i);
	    	Collections.sort(b);
	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
	    }

		
	
	
    static class FastScanner{
    	
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
    	String next() {
    		while(!st.hasMoreTokens())
    			try {
    				st = new StringTokenizer(br.readLine());
    			} catch (IOException e){
    				e.printStackTrace();
    			}
    		return st.nextToken();
    	}
    	
    	int nextInt() {
    		return Integer.parseInt(next());
    	}
    	
    	int [] readArray(int n) {
    		int [] a = new int[n];
    		for(int i=0; i<n ; i++) a[i] = nextInt();
    		return a;
    	}
    	
    	long nextLong() {
    		return Long.parseLong(next());
    	}
    	
    	
    }
    
   
}

