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.
// package testing;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main {
	static public class Pair{
		int a,b;
		public Pair(int a,int b) {
			this.a = a;
			this.b = b;
		}
	}
	static class cmp implements Comparator<Pair>{
		public int compare(Pair p,Pair q) {
			if(p.a<q.a) {
				return -1;
			}
			return 1;
		}
	}
	
	public static void main(String[] args) {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int test = 1;
		test = fs.nextInt();
		outer:
		while(test-->0) {	 
		  int n,m;
		  n = fs.nextInt();
		  m = fs.nextInt();
		  int arr[] = fs.readArray(n);
		  int brr[] = fs.readArray(n);
		  ArrayList<Integer>one = new ArrayList<>();
		  ArrayList<Integer>two = new ArrayList<>();
		  for(int i=0;i<n;i++) {
			  if(brr[i]==1) {
				  one.add(arr[i]);
			  }
			  else {
				  two.add(arr[i]);
			  }
		  }
		  long presum1[] = new long[one.size()+1];
		  long presum2[] = new long[two.size()+1];
		  Collections.sort(one);
		  Collections.sort(two);
		  Collections.reverse(two);
		  Collections.reverse(one);
		  if(one.size()>=1) {
			  presum1[0] = one.get(0);
		  }
		  if(two.size()>=1) {
			  presum2[0] = two.get(0);
		  }
		  for(int i=1;i<one.size();i++) {
			  presum1[i] = presum1[i-1]+one.get(i);
		  }
		  for(int i=1;i<two.size();i++) {
			  presum2[i] = presum2[i-1]+two.get(i);
		  }
		  long sum = 0;
		  long ans = 1000000000000L;
		  for(int i=0;i<one.size();i++) {
			  sum+=one.get(i);
			  if(sum>=m) {
				  ans = Math.min(ans, (i+1));
			  }
			  int l = 0;
			  int r = two.size()-1;
			  int k = -1;
			  while(l<=r) {
				  int mid = (l+r)/2;
				  if(presum2[mid]+sum>=m) {
					  k = mid;
					  r = mid-1;
				  }
				  else {
					  l = mid+1;
				  }
			  }
			  if(k!=-1) {
				  ans = Math.min(ans, (k+1)*2+(i+1));
			  }
		  }
		  sum  = 0;
		  for(int i=0;i<two.size();i++) {
			  sum+=two.get(i);
			  if(sum>=m) {
				  ans = Math.min(ans, (i+1)*2);
			  }
			  int l = 0;
			  int r = one.size()-1;
			  int k = -1;
			  while(l<=r) {
				  int mid = (l+r)/2;
				  if(presum1[mid]+sum>=m) {
					  k = mid;
					  r = mid-1;
				  }
				  else {
					  l = mid+1;
				  }
			  }
			  if(k!=-1) {
				  ans = Math.min(ans, (k+1)+(i+1)*2);
			  }
		  }
		  if(ans==1000000000000L) {
			  ans = -1;
		  }
		  out.println(ans);
		 
    	}
		out.close();
	}

	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());
		}
	}

}
