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.*;

public class codeChef {
    public static void main(String[] args)
    {
        Scanner r = new Scanner(System.in);

        int test = r.nextInt();
        while(test-->0)
        {
            int apps = r.nextInt();
            int mem = r.nextInt();

            ArrayList<Integer> list1 = new ArrayList<>();
            ArrayList<Integer> list2 = new ArrayList<>();
            int[] arr1 = new int[apps];
            int[] arr2 = new int[apps];

            for(int i=0;i<apps;i++)
            {
                arr1[i]=r.nextInt();
            }

            for(int i=0;i<apps;i++)
            {
                arr2[i]=r.nextInt();
            }

            for(int i=0;i<apps;i++)
            {
                int num1 = arr1[i];
                int num2 = arr2[i];

                if(num2==1)
                {
                    list1.add(num1);
                }
                else
                {
                    list2.add(num1);
                }
            }
            Collections.sort(list1);
            Collections.sort(list2);

            int len1 = list1.size();
            int len2 = list2.size();
            len1--;
            len2--;
            boolean bool1 = false;

            int count1 = 0;

            me1: while(mem>0)
            {
                if(len1<0 && len2<0)
                {
                    break me1;
                }

                int first = 0;
                int sec = 0;

                if(len1>=0)
                {
                    first += list1.get(len1);
                    if(len1>0)
                    {
                        first+=list1.get(len1-1);
                    }
                }
                if(len2>-1)
                {
                    sec=list2.get(len2);
                }

                if(first>sec)
                {
                    mem-=first;
                    len1--;
                    count1++;
                    if(len1>=0)
                    {
                        len1--;
                        count1++;
                    }
                }
                else
                {
                    mem-=sec;
                    len2--;
                    count1+=2;
                }
            }

            if(mem<=0)
            {
                int index = 0;
                me2: while(mem<0 && index<list1.size() )
                {
                    if(mem+list1.get(index)<=0)
                    {
                        mem+=list1.get(index++);
                        count1--;
                    }else
                    {
                        break me2;
                    }
                }
                bool1=true;
            }
            if(bool1)
            {
                System.out.println(Math.max(1,count1));
            }
            else
            {
                System.out.println(-1);
            }
        }
    }
}

