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.
//I AM THE CREED
/* //I AM THE CREED
/* package codechef; // don't place package name! */
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.StringTokenizer; 
import java.util.*;
import java.awt.Point;
public class Main{
    static final Random random=new Random();
    public static void main(String[] args) throws IOException 
    { 
        Scanner input=new Scanner(System.in);
        input.nextInt();
        while(input.hasNext()){
            int n=input.nextInt();
            long m=input.nextLong();
            ArrayList<Long> ones=new ArrayList<>();
            ArrayList<Long> two=new ArrayList<>();
            long[] mem=new long[n];
            for(int i=0;i<n;i++){
                mem[i]=input.nextLong();
            }
            for(int i=0;i<n;i++){
                int conv=input.nextInt();
                if(conv==1){
                    ones.add(mem[i]);
                }
                if(conv==2){
                    two.add(mem[i]);
                }
            }
            Collections.sort(ones);
            Collections.sort(two);
            int ans=Integer.MAX_VALUE;
            for(int i=two.size()-2;i>=0;i--){
                two.set(i, two.get(i)+two.get(i+1));
                if(two.get(i)>=m){
                    ans=Math.min(ans, 2*(two.size()-i));
                }
            }
            for(int i=ones.size()-2;i>=0;i--){
                ones.set(i, ones.get(i)+ones.get(i+1));
            }
            for(int i=ones.size();i>=0;i--){
                long rem=m-(i<ones.size()?ones.get(i):0);
                int high=two.size()-1;
                int low=0;
                if(rem<=0){
                    ans=Math.min(ans, ones.size()-i);
                    continue;
                }
                if(two.size()==0){
                    continue;
                }
                while(high>low){
                    int mid=low+((high-low)/2);
                    if(high==low+1){
                        mid=high;
                    }
                    if(two.get(mid)>=rem){
                        low=mid;
                        continue;
                    }
                    high=mid-1;
                    
                }
                if(high>=0 && high<ones.size() && rem>two.get(high)){
                    continue;
                }
                ans=Math.min(ans, ones.size()-i+2*(two.size()-high));
            }
            System.out.println(ans==Integer.MAX_VALUE?-1:ans);
            
        }
        
    }
    static boolean isPrime(int n)
    {
 
        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;
 
        // Check if number is 2
        else if (n == 2)
            return true;
 
        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;
 
        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) 
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
     
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
     
    //Credits to SecondThread(https://codeforces.com/profile/SecondThread) for this tempelate
    static void ruffleSort(int [] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n);
			int temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
    //Credits to SecondThread(https://codeforces.com/profile/SecondThread) for this tempelate.
    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());
		}
	}
    
    
}