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.math.BigInteger;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            BigInteger m = BigInteger.valueOf(sc.nextLong());
            long[] a = new long[n];
            long[] b = new long[n];
            for (int j = 0; j < n; j++) {
                a[j] = sc.nextInt();
            }
            List<Long> imp = new ArrayList<>();
            List<Long> unimp = new ArrayList<>();
            for (int j = 0; j < n; j++) {
                b[j] = sc.nextInt();
                if (b[j] == 1) {
                    unimp.add(a[j]);
                } else {
                    imp.add(a[j]);
                }
            }

            imp.sort(Collections.reverseOrder());
            unimp.sort(Collections.reverseOrder());

            BigInteger free = BigInteger.ZERO;
            int otvet = 0;
            while (free.compareTo(m) < 0 && unimp.size() >= 3 && !imp.isEmpty()) {
                BigInteger runningSum = BigInteger.ZERO;
                int j = 0;
                for (; j < unimp.size(); j++) {
                    BigInteger newSum = runningSum.add(BigInteger.valueOf(unimp.get(j)));
                    if (newSum.compareTo(BigInteger.valueOf(imp.get(0))) > 0
                            || newSum.compareTo(m) >= 0) {
                        break;
                    }
                    runningSum = newSum;
                }
                if (j > 1) {
                    free = free.add(BigInteger.valueOf(imp.remove(0)));
                    otvet += 2;
                } else {
//                    if (free.add(BigInteger.valueOf(unimp.get(0))).compareTo(m) < 0 && free.add(BigInteger.valueOf(imp.get(0))).compareTo(m) >= 0) {
//                        free = free.add(BigInteger.valueOf(imp.remove(0)));
//                        otvet += 2;
//                        break;
//                    } else {
                    free = free.add(BigInteger.valueOf(unimp.remove(0)));
//                    }
                    otvet++;
                }
            }

            if (free.compareTo(m) >= 0) {
                System.out.println(otvet);
                continue;
            }

            int unimpIndex = 0;
            while (free.compareTo(m) < 0 && unimpIndex < unimp.size()) {
                free = free.add(BigInteger.valueOf(unimp.get(unimpIndex++)));
                otvet++;
            }
            while (free.compareTo(m) < 0 && !imp.isEmpty()) {
                free = free.add(BigInteger.valueOf(imp.remove(0)));
                otvet += 2;
            }
            while (unimpIndex > 0 && free.subtract(BigInteger.valueOf(unimp.get(unimpIndex - 1))).compareTo(m) >= 0) {
                free = free.subtract(BigInteger.valueOf(unimp.get(--unimpIndex)));
                otvet--;
            }

            if (free.compareTo(m) < 0) {
                System.out.println(-1);
            } else {
                System.out.println(otvet);
            }
        }
    }
}
