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.Collections;
import java.util.Comparator;
import java.util.StringTokenizer;

public class CleaningThePhone {
	static class FastScanner 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastScanner() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 

		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	} 
	 
	
	public static void main(String[] args) {
		FastScanner f=new FastScanner();
		int t=f.nextInt();
		while(t-->0) {
			int n=f.nextInt();
			long k=f.nextLong();
			long w[]=new long[n];
			long v[]=new long[n];
			long sum=0;
			for (int i = 0; i < w.length; i++) {
				w[i]=f.nextLong();
				sum+=w[i];
			}
			for (int i = 0; i < v.length; i++) {
				v[i]=f.nextLong();
			}
			
			ArrayList<Long> one=new ArrayList<>();
			ArrayList<Long> two=new ArrayList<>();
			for(int i=0;i<n;i++) {
				if(v[i]==1) one.add(w[i]);
				else two.add(w[i]);
			}
			
			Collections.sort(one);
			Collections.reverse(one);
			Collections.sort(two);
			Collections.reverse(two);
			
			answer(one,two,k,sum);
			
		}
	}
	
	static void answer(ArrayList<Long> one,ArrayList<Long> two,long k,long sum) {
		if(sum<k) {
			System.out.println(-1);
			return;
		}
		long ans=0;
		long tempk=0;
		boolean answerfound=false;
		for(int i=0;i<one.size();i++) {
			if(tempk<k) {
				tempk+=one.get(i);
				ans++;
			}else {
				answerfound=true;
				break;
			}
		}
		
		if(answerfound==true) {
			System.out.println(ans);
			return;
		}
		
		for(int i=0;i<two.size();i++) {
			if(tempk<k) {
				tempk+=two.get(i);
				ans+=2;
			}else if(tempk==k) {
				answerfound=true;
				break;
			}
		}
		
		if(answerfound==true) {
			System.out.println(ans);
			return;
		}
		
		for(int i=one.size()-1;i>=0;i--) {
			long curr=one.get(i);
			if(tempk-curr>=k) {
				tempk=tempk-curr;
				ans--;
			}else {
				answerfound=true;
				break;
			}
		}
		
		if(answerfound==true) {
			System.out.println(ans);
			return;
		}else {
			System.out.println(-1);
		}
		
	}

}
