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.

// D
import java.util.*;
// import java.lang.*;
import java.io.*;

//           THIS TEMPLATE MADE BY AKSH BANSAL.

public class Solution {
    public static void main(String[] args) {
 
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt();
        while (t-- > 0) {
            int n = scanner.nextInt();
            int m = scanner.nextInt();
            int[][] rank = new int[n][2];
            long tot = 0;
            for(int i=0;i<n;i++)
                tot+=rank[i][1] = scanner.nextInt();
            for(int i=0;i<n;i++)
                rank[i][0] = scanner.nextInt();
            if(tot<m){System.out.println("-1");continue;}
            long cur = 0;
            Arrays.sort(rank, new Comparator<int[]>() {
                @Override
                public int compare(int[] o1, int[] o2) {
                    if(o1[0]==o2[0])
                    {
                        return o1[0]==1?o1[1]-o2[1]:o2[1]-o1[1];
                    }
                    return o1[0]-o2[0];
                }
            });
 
            int l = 0;
            int cost = 0;
            int ans = Integer.MAX_VALUE;
            for(int i=0;i<n;i++)
            {
                cur+=rank[i][1];
                cost+=rank[i][0];
                while(cur>=m){ans = Math.min(ans,cost);cur-=rank[l][1];cost-=rank[l++][0];}
            }
 
            System.out.println(ans);
 
 
        }
    }
    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;
        }
    }

    // public static void main(String[] args) throws IOException {
    //     FastReader sc = new FastReader();
    //     PrintWriter out = new PrintWriter(System.out);
    //     // ________________________________

    //     int t = sc.nextInt();
    //     // StringBuilder output = new StringBuilder();

    //     while (t-- > 0) {
    //         int n = sc.nextInt();
    //         int m = sc.nextInt();
    //         // Integer[][] arr = new Integer[n][2];
    //         int[][] arr = new int[n][2];
    //         int sum = 0;
    //         for(int i=0;i<n;i++){
    //             arr[i][0] = sc.nextInt();
    //             sum+=arr[i][0];
    //         }
    //         for(int i=0;i<n;i++){
    //             arr[i][1] = sc.nextInt();
    //         }
    //         if(sum<m){
    //             // output.append("-1").append("\n");
    //             out.println(-1);
    //             continue;
    //         }
    //         // output.append(solver(arr, n, m)).append("\n");
    //         out.println(solver(arr, n, m));
    //     }

    //     // out.println(output);
    //     // _______________________________

    //     // int n = sc.nextInt();
    //     // out.println(solver());
    //     // ________________________________
    //     out.flush();
    // }

    public static long solver(int[][] arr,int n,int m) {            
        // Arrays.sort(arr,(a,b)->
        //     a[1]==b[1]?
        //         a[1]==1?
        //             a[0]-b[0]
        //         :
        //             b[0]-a[0]
        //     :
        //         a[1]-b[1]
        // );
        Arrays.sort(arr, new Comparator<int[]>() {
                @Override
                public int compare(int[] o1, int[] o2) {
                    if(o1[0]==o2[0])
                    {
                        return o1[0]==1?o1[1]-o2[1]:o2[1]-o1[1];
                    }
                    return o1[0]-o2[0];
                }
            }
        );
       // for(Integer[] i:arr)System.out.print("__"+ i[0]);

        int res=Integer.MAX_VALUE;
        int cost = 0;
        long cur = 0;
        int l = 0;
        for(int i=0;i<n;i++){
                cur+=arr[i][0];
                cost+=arr[i][1];
            while(cur>=m){
                res = Math.min(cost, res);
                cur-=arr[l][0];
                cost-=arr[l++][1];
            }
        }
        return res;
    }
}