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.util.*; 
import java.math.*; 

public class Main {
	
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=f.nextInt();
//		int t=1;
		while(t>0) {
			t--;
			int n=f.nextInt();
			int m=f.nextInt();
			int[] a=f.readArray(n);
			int[] b=f.readArray(n);
			int sum=0;
			int count=0;
			for(int i=0;i<n;i++) {
				sum+=a[i];
				if(b[i]==1) {
					count++;
				}
					
			}
			if(sum<m) {
				System.out.println(-1);
			}
			else {
				int[] one=new int[count];
				int[] two=new int[n-count];
				
				int startone=0;
				int starttwo=0;
				for(int i=0;i<n;i++) {
					if(b[i]==1) {
						one[startone]=a[i];
						startone++;
					}
					else {
						two[starttwo]=a[i];
						starttwo++;
					}
				}
				sort(one);
				sort(two);
				for(int i=one.length-2;i>-1;i--) {
					one[i]=one[i]+one[i+1];
//					System.out.println(one[i]);
				}
				for(int i=two.length-2;i>-1;i--) {
					two[i]=two[i]+two[i+1];
//					System.out.println(two[i]);
				}
				int end=two.length-1;
				int ans=1000000000;
//				if(one.length==0) {
//					ans=Math.min(ans, 2*);
//				}
				for(int i=0;i<one.length;i++) {
					int rest=m-one[i];
//					System.out.println(rest+" "+ans+" "+end);
					if(rest<0) {
						ans=Math.min(ans,one.length-i);
						continue;
					}
					else {
						if(rest==0) {
							ans=Math.min(ans,one.length-i);
							continue;
						}
						while(end>-1 && rest>two[end]) {
							end--;
						}
						if(end<0) {
							break;
						}
						ans=Math.min(ans, one.length-i+2*(two.length-end));
						
						
					}
					
				}
//				System.out.println(two[0]);
				for(int i=0;i<two.length;i++) {
					if(two[i]>=m) {
						ans=Math.min(ans, 2*(two.length));
					}
				}
				System.out.println(ans);
			}
		}
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
	
	static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
    
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;

    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 