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 D{
	static class Pair{
		int m;
		int b;
		Pair(int mm, int bb){
			b = bb;
			m = mm;
		}
		Pair(int mm){
			m = mm;
		}
		
		public String toString(){
			return new Formatter().format("(%d,%d)",m,b).toString();
		}
	}
	
	public static void main(String[] args){
		MyScanner in = new MyScanner();
		StringBuilder res = new StringBuilder();
		int t = in.nextInt();
		Integer[] a = new Integer[200000];
		Integer[] b = new Integer[200000];
		int[] ms = new int[200000];
	//	Comparator<Pair> comp = (x,y)-> (x.b==y.b ? -Integer.compare(x.m,y.m): Integer.compare(x.b,y.b));
		o:while(t-- > 0){
			int n = in.nextInt();
			long m = in.nextInt();
			//Pair[] app = new Pair[n];
			for(int i = 0;i<n;++i){
				ms[i] = in.nextInt();
			}
			int acount =0;
			int bcount =0;
			long asum = 0;
			int mm = 0;
			for(int i = 0;i<n;++i){
				mm = in.nextInt();
				if(mm==1){
					a[acount++] = ms[i];
					asum += ms[i];
				} else{
					b[bcount++] = ms[i];
				}
			}
			Arrays.sort(a,0,acount,(x,y)->Integer.compare(y,x));
			Arrays.sort(b,0,bcount,(x,y)->Integer.compare(y,x));
			TreeMap<Long,Integer> map = new TreeMap<>();
			long sum = 0;
			map.put(0l,0);
			for(int i = 0;i<bcount;++i){
				map.put(sum +b[i],(i+1)*2);
				sum += b[i];
			}
			sum = 0;
			int best = 0;
			if(map.containsKey(m)){
				best = map.get(m);
			} else if(map.higherKey(m) != null){
				best = map.get(map.higherKey(m));
			} else{
				best = Integer.MAX_VALUE;
			}
			int r = 0;
			for(int i = 0;i<acount;++i){
				sum += a[i];
				r++;
				if(map.containsKey(m-sum)){
					best = Math.min(best,map.get(m-sum) + r);
				} else if(map.higherKey(m-sum) != null){
					best = Math.min(best,map.get(map.higherKey(m-sum))+r);
				}
			}
			res.append((best == Integer.MAX_VALUE ? -1 : best) + "\n");
		}
		
		System.out.println(res.toString());
		
	}
	
	static class MyScanner{
		BufferedReader bf;
		StringTokenizer st;
		
		MyScanner(){
			bf = new BufferedReader(new InputStreamReader(System.in));
		}
		
		public String nextToken(){
			try{
				while(st == null || !st.hasMoreTokens()){
					st = new StringTokenizer(bf.readLine());
				}
				return st.nextToken();
			} catch(Exception e){
				System.out.println("Problems");
			}
			return "";
		}
		
		public Integer nextInt(){
			return Integer.valueOf(nextToken());
		}
	}
}

