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.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
	/**
	 * 选够价值是m的物品且Bi最小
	 *
	 * bi可以被二分
	 *
	 * 按b排序
	 *
	 * 我去枚举背包容量b 使得背包价值>=m
	 *
	 * 01背包 n^2 GG
	 *
	 * 考虑贪心
	 *
	 * 在最优解中
	 *
	 * 如果有m值比较大且b相同 那我一定选m大的
	 *
	 * 我把所有的物品分成两组 一组是b=1 一组是b=2 ,从小到大排序
	 * 那么我一定先选b=1
	 *
	 * 可不可以直接贪心
	 *
	 * 按m/b排序？
	 *
	 * 10000/2  5001/1
	 * 100/2
	 *
	 * 答案如果是要5002 那我算出来的是3  实际上是2
	 *
	 * 排完序以后做O1背包？
	 *
	 *
	 *
	 * @param args
	 */
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int T = scan.nextInt();
		while(T-- > 0) {
			int n = scan.nextInt();
			int m = scan.nextInt();
			List<Integer> cnt1 = new ArrayList<>();
			List<Integer> cnt2 = new ArrayList<>();
			int[] cnt = new int[n];
			for (int i = 0; i < n; i++) cnt[i] = scan.nextInt();
			for (int i = 0; i < n; i++) {
				int b = scan.nextInt();
				if (b == 1) cnt1.add(cnt[i]);
				else cnt2.add(cnt[i]);
			}
			Collections.sort(cnt1, Collections.reverseOrder());
			Collections.sort(cnt2, Collections.reverseOrder());
			int[] sum1 = new int[cnt1.size()+1];
			int[] sum2 = new int[cnt2.size()+1];
			int n1 = cnt1.size();
			int n2 = cnt2.size();
			for(int i=1;i<=cnt1.size();i++)
				sum1[i] = cnt1.get(i-1) + sum1[i-1];
			for(int i=1;i<=cnt2.size();i++)
				sum2[i] = cnt2.get(i-1) + sum2[i-1];
			int res = 0x3f3f3f3f;
			for(int i=0;i<=n1;i++){
				int cm = m - sum1[i];
				if(cm<=0){
					res = Math.min(res,i);
					continue;
				}
				int l = 0,r = n2;
				while( l<r){
					int mid = l + r >> 1;
					if(sum2[mid]>=cm){
						r = mid;
					}else{
						l = mid + 1;
					}
				}
				if(sum2[l]>=cm)
				res = Math.min(res,i+l*2);
			}
			if(res == 0x3f3f3f3f){
				System.out.println(-1);
			}else{
				System.out.println(res);
			}
		}
	}
}

