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 CleaningthePhone {

	// https://codeforces.com/contest/1475/problem/D
	
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("CleaningthePhone"));

		int t = Integer.parseInt(in.readLine());
		while (t-->0) {
			StringTokenizer st = new StringTokenizer(in.readLine());
			int n = Integer.parseInt(st.nextToken());
			int m = Integer.parseInt(st.nextToken());
			long[] arr = new long[n];
			st = new StringTokenizer(in.readLine());
			for (int i=0; i<n; i++) {
				arr[i] = Integer.parseInt(st.nextToken());
			}
			st = new StringTokenizer(in.readLine());
			int[] b = new int[n];
			for (int i=0; i<n; i++ ) {
				b[i] = Integer.parseInt(st.nextToken());
			}
			
			
			ArrayList<Long> ones = new ArrayList<>(), twos = new ArrayList<>();
			for (int i=0; i<n; i++) {
				if (b[i] == 1) {
					ones.add(arr[i]);
				}
				else {
					twos.add(arr[i]);
				}
			}
			Collections.sort(ones);
			Collections.sort(twos);
			
			int p1=ones.size()-1;
			int p2=twos.size()-1;
			long total = 0;
			int bad=0;
			
			while (total < m && p1 >= 0 && p2 >= 0) {
				if (total + ones.get(p1) >= m) {
					total += ones.get(p1);
					p1--;
					bad++;
					break;
				}
				if (total + twos.get(p2) >= m) {
					total += twos.get(p2);
					p2--;
					bad += 2;
					break;
				}
				if (p1-1 >= 0) {
					if (ones.get(p1) + ones.get(p1-1) >= twos.get(p2)) {
						total += ones.get(p1);
						p1--;
						bad++;
					}
					else {
						total += twos.get(p2);
						p2--;
						bad += 2;
					}
				}
				else {
					if (ones.get(p1) >= twos.get(p2)) {
						total += ones.get(p1);
						p1--;
						bad++;
					}
					else {
						total += twos.get(p2);
						p2--;
						bad += 2;
					}
				}
			}
			
			while (total < m && p1 >= 0) {
				total += ones.get(p1);
				p1--;
				bad++;
			}
			
			while (total < m && p2 >= 0) {
				total += twos.get(p2);
				p2--;
				bad += 2;
			}
			
			if (total >= m) {
				System.out.println(bad);
			}
			else {
				System.out.println(-1);
			}
			
		}
	}
}