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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;

/**
 * @author Peter Li
 * @version 01/29/2021
 */
public class Cleaning2 {
    public static void main(String[] args) throws IOException {
        int d = ni();
        while (d-- > 0) {
            int n = ni(), m = ni();
            ArrayList<Integer> one = new ArrayList<>(), two = new ArrayList<>();
            int[] mem = na(n);
            for (int i = 0; i < n; i++) {
                int cost = ni();
                if (cost == 1) {
                    one.add(mem[i]);
                } else {
                    two.add(mem[i]);
                }
            }
            Collections.sort(one);
            Collections.reverse(one);
            Collections.sort(two);
            Collections.reverse(two);
            long[] o = new long[one.size() + 1], t = new long[two.size() + 1];
            for (int i = 1; i <= one.size(); i++) {
                o[i] = o[i - 1] + one.get(i - 1);
            }
            for (int i = 1; i <= two.size(); i++) {
                t[i] = t[i - 1] + two.get(i - 1);
            }
            int oneCount = 0, twoCount = two.size();
            while (oneCount < o.length && o[oneCount] + t[t.length - 1] < m) {
                oneCount += 1;
            }
            if (oneCount == o.length) {
                System.out.println(-1);
                continue;
            }
            int ans = oneCount + twoCount * 2;
            while (oneCount <= one.size()) {
                while (twoCount > 0 && o[oneCount] + t[twoCount - 1] >= m) {
                    twoCount--;
                }
                ans = Math.min(ans, oneCount + twoCount * 2);
                oneCount++;
            }
            System.out.println(ans);
        }
    }
    
    private final static BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 1 << 15);
    private static String[] line = new String[0];
    private static int lineIndex = 0;
    
    private static void checkLine() {
        if (lineIndex == line.length) {
            try {
                line = br.readLine().split(" ");
            } catch (IOException e) {
                throw new RuntimeException("Unexpected input");
            }
            lineIndex = 0;
        }
    }
    
    private static String ns() {
        checkLine();
        return line[lineIndex++];
    }
    
    private static long nl() {
        return Long.parseLong(ns());
    }
    
    private static int ni() {
        return (int) nl();
    }
    
    private static double nd() {
        return Double.parseDouble(ns());
    }
    
    private static int[] na(int n) {
        int[] ans = new int[n];
        for (int i = 0; i < n; i++) {
            ans[i] = ni();
        }
        return ans;
    }
    
    private static String[] nsa(int n) {
        String[] ans = new String[n];
        for (int i = 0; i < n; i++) {
            ans[i] = ns();
        }
        return ans;
    }
}
