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 Solution{
	static PrintWriter out = new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		
		String[] input = br.readLine().trim().split(" ");
		
		int numTestCases = Integer.parseInt(input[0]);
		
		
		while(numTestCases-- > 0){
		    input = br.readLine().trim().split(" ");
    		int n = Integer.parseInt(input[0]);
    		int m = Integer.parseInt(input[1]);
    		
    		int[] a = new int[n];
    		int[] b = new int[n];
    		input = br.readLine().trim().split(" ");
    		for(int i = 0; i < n; i++){
    		    a[i] = Integer.parseInt(input[i]);
    		}
    		input = br.readLine().trim().split(" ");
    		for(int i = 0; i < n; i++){
    		    b[i] = Integer.parseInt(input[i]);
    		}
    		out.println(minConveniencePoints(a, b, m));
		}
		
		out.flush();
		out.close();
	}
	public static int minConveniencePoints(int[] arr, int[] type, int m)
	{
	    long sum = 0;
	    for(int i = 0; i < arr.length; i++){
	        sum += arr[i];
	    }
	    if(sum < m){
	        return -1;
	    }
	    ArrayList<Integer> importantAppSizes = new ArrayList<>();
	    ArrayList<Integer> regularAppSizes = new ArrayList<>();
	    
	    int n = arr.length;
	    for(int i = 0; i < n; i++){
	        if(type[i] == 1)
	        {
	            regularAppSizes.add(arr[i]);
	        }
	        else{
	            importantAppSizes.add(arr[i]);
	        }
	    }
	    Collections.sort(regularAppSizes, Collections.reverseOrder());
	    Collections.sort(importantAppSizes, Collections.reverseOrder());
	    
	    
	    sum = 0;
	    long[] prefixSum = new long[regularAppSizes.size()];
	    if(0 < regularAppSizes.size()){
	        prefixSum[0] = regularAppSizes.get(0);
	    }
	    for(int i = 1; i < regularAppSizes.size(); i++){
	        prefixSum[i] = prefixSum[i - 1] + regularAppSizes.get(i);
	    }
	    int ans = binarySearch(prefixSum, m);
	    if(ans == -1){
	        ans = Integer.MAX_VALUE;
	    }
	    else{
	        ans++;
	    }
	    for(int i = 0; i < importantAppSizes.size(); i++){
	        sum += importantAppSizes.get(i);
	        if(sum >= m)
	        {
	            ans = Math.min(ans, ((i + 1) * 2));
	        }
	        else{
    	        long requiredSize = m - sum;
    	        int index = binarySearch(prefixSum, requiredSize);
    	        if(index != -1)
    	        {
    	            int option = ((i + 1) * 2) + (index + 1); 
    	            ans = Math.min(ans, option);
    	        }
	        }
	    }
	    return ans;
	}
	public static int binarySearch(long[] arr, long x)
	{
	    int ans = -1;
	    int start = 0, end = arr.length - 1;
	    while(start <= end){
	        int mid = (start + end) / 2;
	        if(arr[mid] >= x)
	        {
	            ans = mid;
	            end = mid - 1;
	        }
	        else{
	            start = mid + 1;
	        }
	    }
	    return ans;
	}
}