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 Solution implements Runnable {
    FastScanner sc;
    PrintWriter pw;
    
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) throws Exception
    {
        new Thread(null,new Solution(),"codeforces",1<<28).start();
    }
    public void run()
    {
        sc=new FastScanner();
        pw=new PrintWriter(System.out);
        try{
        solve();}
        catch(Exception e)
        {
            pw.println(e);
        }
        pw.flush();
        pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    public long ppow(long a,long b,long mod)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1L)
            tmp*=a;
            a*=a;
            a%=mod;
            tmp%=mod;
            b>>=1;
        }
        return (tmp*a)%mod;
    }
    public long pow(long a,long b)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1)
            tmp*=a;
            a*=a;
            b>>=1;
        }
        return (tmp*a);
    }
    public  int gcd(int x,int y)
    {
        return y==0?x:gcd(y,x%y);
    }
   
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////
    public void solve() throws Exception{
       int t=sc.ni();
       while(t-->0){
           int n=sc.ni();
           long m=sc.ni();
           long[] arr=new long[n];
           for(int i=0;i<n;i++)
           arr[i]=sc.nlo();
           ArrayList<Long> list=new ArrayList();
           ArrayList<Long> list1=new ArrayList();
           for(int i=0;i<n;i++){
               int x=sc.ni();
               if(x==1)
               list.add(arr[i]);
               else
               list1.add(arr[i]);
           }
           Collections.sort(list,Collections.reverseOrder());
           Collections.sort(list1,Collections.reverseOrder());
           long pre=0;
           for(int i=0;i<list.size();i++){
               pre+=list.get(i);
               list.set(i,pre);
              // pw.println(pre);
           }
           pre=0;
           for(int i=0;i<list1.size();i++){
               pre+=list1.get(i);
               list1.set(i,pre);
           }
           int ans=Integer.MAX_VALUE;
           int tmp=bs(list1,m);
           if(tmp!=-1)
           ans=2*(tmp+1);
           for(int i=0;i<list.size();i++)
           {
               long val=list.get(i);
                if(val>=m){
                    ans=Math.min(ans,i+1);
                    break;
                }
                int p=bs(list1,m-val);
                //pw.println(p+" "+i);
                if(p!=-1)
                    ans=Math.min(ans,i+1+2*(p+1));
           }
           if(ans==Integer.MAX_VALUE)
           ans=-1;
           pw.println(ans);
        }
    }
    public int bs(ArrayList<Long> list,long val){
        int l=0,ans=-1,r=list.size()-1;
        while(l<=r){
            int m=(l+r)/2;
            if(list.get(m)>=val){
                r=m-1;
                ans=m;
            }
            else
            l=m+1;
        }
        return ans;
    }
}