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.
///package div3;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class board
{
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }


    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // method to return LCM of two numbers
    static long lcm(long a, long b)
    {
        return (a / gcd(a, b)) * b;
    }
    static void sieveOfEratosthenes(int n)
    {
        // Create a boolean array "prime[0..n]" and initialize
        // all entries it as true. A value in prime[i] will
        // finally be false if i is Not a prime, else true.

        for(int i=0;i<n;i++)
            prime[i] = true;

        for(int p = 2; p*p <=n; p++)
        {
            // If prime[p] is not changed, then it is a prime
            if(prime[p] == true)
            {
                // Update all multiples of p
                for(int i = p*p; i <= n; i += p)
                    prime[i] = false;
            }
        }


    }
    static boolean[] prime;

    static int UpperBound(long a[], long x) {// x is the key or target value
        int lo = 0, hi = a.length;
        int ans = hi+1;
        while(lo < hi) {
            int mid = (lo + hi)/2;
            if(x<a[mid])
                hi = mid;
            else if(x == a[mid]){
                ans = mid;
                lo = mid + 1;
            }
            else lo = mid+1;
        }
        return Math.min(ans,lo);
    }
    public static void main(String[] args)
    {
        FastReader sc=new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = sc.nextInt();

        while(t-->0) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int[] a = new int[n];
            int[] b = new int[n];
            for(int i=0;i<n;i++){
                a[i] = sc.nextInt();
            }

            List<Integer> ones = new ArrayList<>();
            List<Integer> twos = new ArrayList<>();
            for(int i=0;i<n;i++){
                b[i] = sc.nextInt();
                if(b[i] == 1)ones.add(a[i]);
                if(b[i] == 2)twos.add(a[i]);
            }
            if(twos.isEmpty()){
                Collections.sort(ones,Collections.reverseOrder());
                int pos = 0;
                int req = m;
                for(int i=0;i<n;i++){
                    if(req > ones.get(i)){
                        req -= ones.get(i);
                    }
                    else{
                        req = 0;
                        pos = i+1;
                        break;
                    }
                }
                if(req == 0) out.println(pos);
                else out.println(-1);
                continue;
            }
            Collections.sort(ones);
            Collections.sort(twos);

            long[] suf_sum = new long[twos.size()];
            suf_sum[twos.size()-1] = twos.get(twos.size()-1);
            for(int i=twos.size()-2;i>=0;i--){
                suf_sum[i] += suf_sum[i+1]+twos.get(i);
            }
            Arrays.sort(suf_sum);
            long one_sum = 0;
            int max = (int)(1e9);
            int min = (suf_sum[suf_sum.length-1]<m)?max:UpperBound(suf_sum,m);
            long ans = (min+1)*2;
            //out.println(ans);
            for(int i=ones.size()-1;i>=0;i--){
                one_sum += ones.get(i);
                long req = m-one_sum;
                long get = (suf_sum[suf_sum.length-1]<req)?max:UpperBound(suf_sum,req);

                //out.println(one_sum+" "+req+" "+get);
                if(req <= 0) get = -1;
                ans = Math.min(ans,ones.size()-i+2*(get+1));

            }
            out.println((ans>=max)?-1:ans);

        }
        out.close();
    }


}