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.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        //BufferedReader f = new BufferedReader(new FileReader("uva.in"));
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int t = Integer.parseInt(f.readLine());
        while(t-- > 0) {
            StringTokenizer st = new StringTokenizer(f.readLine());
            int n = Integer.parseInt(st.nextToken());
            int m = Integer.parseInt(st.nextToken());
            st = new StringTokenizer(f.readLine());
            int[] a = new int[n];
            for(int i = 0; i < n; i++) {
                a[i] = Integer.parseInt(st.nextToken());
            }
            st = new StringTokenizer(f.readLine());
            int[] b = new int[n];
            int ones = 0;
            int twos = 0;
            for(int i = 0; i < n; i++) {
                b[i] = Integer.parseInt(st.nextToken());
                if(b[i] == 1) {
                    ones++;
                } else {
                    twos++;
                }
            }
            int high = ones+2*twos;
            int[] regular = new int[ones];
            int[] important = new int[twos];
            for(int i = 0; i < n; i++) {
                if(b[i] == 1) {
                    regular[--ones] = a[i];
                } else {
                    important[--twos] = a[i];
                }
            }
            Arrays.sort(regular);
            Arrays.sort(important);
            long[] prefixSum1 = new long[regular.length+1];
            for(int i = 1; i <= regular.length; i++) {
                prefixSum1[i] = prefixSum1[i-1]+regular[regular.length-i];
            }
            long[] prefixSum2 = new long[important.length+1];
            for(int i = 1; i <= important.length; i++) {
                prefixSum2[i] = prefixSum2[i-1]+important[important.length-i];
            }
            int low = 0;
            int ans = -1;
            while(low <= high) {
                int mid = (low+high)/2;
                boolean valid = false;
                for(int i = 0; i <= Math.min(regular.length, mid); i++) {
                    if(prefixSum1[i]+prefixSum2[Math.min(important.length, (mid-i)/2)] >= m) {
                        valid = true;
                        break;
                    }
                }
                if(valid) {
                    ans = mid;
                    high = mid-1;
                } else {
                    low = mid+1;
                }
            }
            out.println(ans);
        }
        f.close();
        out.close();
    }
}