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) {
		int t = Input.ni();
		while(t--!=0) {
			int n = Input.ni();
			int m = Input.ni();
			int [] a = new int [n+1];
			int [] b = new int [n+1];
			String [] s1 = Input.nline().split(" ");
			String [] s2 = Input.nline().split(" ");
			for(int i=1;i<=n;i++) {
				if(s2[i-1].equals("1")) a[++a[0]] = Integer.parseInt(s1[i-1]);
				else b[++b[0]] = Integer.parseInt(s1[i-1]);
			}
			Arrays.sort(a,1,a[0]+1);
			Arrays.sort(b,1,b[0]+1);
			solve(a,b,m,n);
		}
	}

	private static void solve(int[] a, int[] b, int m, int n) {
		int aa = a[0];
		int bb = b[0];
		int sum = 0;
		double k = b[b[0]]/2;
		while(m>0&&n>0) {
			if(m-a[a[0]]<=0) {
				m -= a[a[0]];
				a[0]--;
				sum++;
				break;
			}
			if(a[a[0]]>k) {
				m -= a[a[0]];
				a[0]--;
				sum++;
			}else {
				m -= b[b[0]];
				b[0]--;
				k = b[b[0]]/2;
				sum+=2;
			}
			n--;
		}
		if(m<0&&b[0]==bb&&a[0]+1==aa) {
			System.out.println(1);
			return ;
		}
		if(a[0]!=aa)a[0]++;
		if(b[0]!=bb)b[0]++;
 
		if(m<0) {
			if(a[a[0]]+m==0) {
				m=0;
				sum--;
			}
			if(b[b[0]]+m==0) {
				m=0;
				sum-=2;
			}
		}
		if(m>0) System.out.println(-1);
		else System.out.println(sum);
	}
}
final class Input {
    static final byte[] b = new byte[1024];
    static int l = 0, pos = 0;
 
    static int readByte() {
        if (l == -1) throw new InputMismatchException();
        if (pos >= l) {
            pos = 0;
            try {
                l = System.in.read(b);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (l <= 0) return -1;
        }
        return b[pos++];
    }
 
    static boolean isSpaceChar(int c) {
        return !(c >= 33 && c <= 126);
    }
 
    static int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b)) ;
        return b;
    }
 
    static String ns() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
 
    static String nline() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!isSpaceChar(b) || b == ' ') {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
 
    static int ni() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') num = (num << 3) + (num << 1) + (b - '0');
            else return minus ? -num : num;
            b = readByte();
        }
    }
 
    static long nl() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') num = num * 10 + (b - '0');
            else return minus ? -num : num;
            b = readByte();
        }
    }
}