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.BigInteger;
import java.util.*;

public class test1{

    public static void main(String[] args) throws IOException{
        long startTime = System.currentTimeMillis();
        Reader sc = new Reader();
		int t = sc.nextInt();
		StringBuilder finalResult= new StringBuilder();
		
        while(t-- > 0){
			int n=sc.nextInt();
			int m=sc.nextInt();
			
			int[] mem=new int[n];
			int[] con=new int[n];
			for(int i=0; i<n; i++)
				mem[i]=sc.nextInt();

			for(int i=0; i<n; i++)
				con[i]=sc.nextInt();

			List<Integer> con1 = new ArrayList<>();
			List<Integer> con2 = new ArrayList<>();
			long sumCon2=0;
			for(int i=0; i<n; i++){
				if(con[i]==1)
					con1.add(mem[i]);
				else{
					con2.add(mem[i]);
					sumCon2 += mem[i];
				}
			}

			Collections.sort(con1);
			Collections.sort(con2);
			Collections.reverse(con1);
			Collections.reverse(con2);

			long temp=sumCon2;
			int cost=2*con2.size();
			long result=Integer.MAX_VALUE;
			for(int i=0, j=con2.size(); i<=con1.size(); i++){
				while(j > 0 && temp-con2.get(j-1) >= m){
					temp -= con2.get(j-1);
					cost-=2;
					j--;
				}

				if(temp >= m){
					result=Math.min(result,cost);
				}

				if(i<con1.size()){
					temp += con1.get(i);
					cost += 1;
				}
			}
			if(result==Integer.MAX_VALUE)
				result=-1;
			finalResult.append(result + "\n");
		}   
		System.out.print(finalResult);
        long endTime = System.currentTimeMillis();
    }

    static class Reader {
	    final private int BUFFER_SIZE = 1 << 16;
	    private DataInputStream din;
	    private byte[] buffer;
	    private int bufferPointer, bytesRead;

	    public Reader() {
	      din = new DataInputStream(System.in);
	      buffer = new byte[BUFFER_SIZE];
	      bufferPointer = bytesRead = 0;
	    }

	    public Reader(String file_name) throws IOException {
	      din = new DataInputStream(new FileInputStream(file_name));
	      buffer = new byte[BUFFER_SIZE];
	      bufferPointer = bytesRead = 0;
	    }

	    public String readLine() throws IOException {
	      byte[] buf = new byte[200002];
	      int cnt = 0, c;
	      while ((c = read()) != -1) {
	        if (c == '\n') break;
	        buf[cnt++] = (byte) c;
	      }
	      return new String(buf, 0, cnt);
	    }

	    public int nextInt() throws IOException {
	      int ret = 0;
	      byte c = read();
	      while (c <= ' ') c = read();
	      boolean neg = (c == '-');
	      if (neg) c = read();
	      do {
	        ret = ret * 10 + c - '0';
	      } while ((c = read()) >= '0' && c <= '9');
	      if (neg) return -ret;
	      return ret;
	    }

	    public long nextLong() throws IOException {
	      long ret = 0;
	      byte c = read();
	      while (c <= ' ') c = read();
	      boolean neg = (c == '-');
	      if (neg) c = read();
	      do {
	        ret = ret * 10 + c - '0';
	      } while ((c = read()) >= '0' && c <= '9');
	      if (neg) return -ret;
	      return ret;
	    }

	    public double nextDouble() throws IOException {
	      double ret = 0, div = 1;
	      byte c = read();
	      while (c <= ' ') c = read();
	      boolean neg = (c == '-');
	      if (neg) c = read();
	      do {
	        ret = ret * 10 + c - '0';
	      } while ((c = read()) >= '0' && c <= '9');
	      if (c == '.')
	        while ((c = read()) >= '0' && c <= '9') ret += (c - '0') / (div *= 10);
	      if (neg) return -ret;
	      return ret;
	    }

	    private void fillBuffer() throws IOException {
	      bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
	      if (bytesRead == -1) buffer[0] = -1;
	    }

	    private byte read() throws IOException {
	      if (bufferPointer == bytesRead) fillBuffer();
	      return buffer[bufferPointer++];
	    }

	    public void close() throws IOException {
	      if (din == null) return;
	      din.close();
	    }
	  }
}

class Pair{
	int x,y;
	Pair(int a, int b){
		x=a;y=b;
	}
}