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 D {

	static long m = (long) (1e9 + 7);

	public static void main(String[] args) throws IOException {
		Scanner scn = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		StringBuilder sb = new StringBuilder();
		int T = scn.ni(), tcs = 0;
		D: while (tcs++ < T) {
			int n = scn.ni();
			long m = scn.nl();
			long a[] = scn.nla(n);
			long b[] = scn.nla(n);
			ArrayList<Long> l1 = new ArrayList<>(), l2 = new ArrayList<>();
			for (int i = 0; i < n; i++) {
				if (b[i] == 1)
					l1.add(a[i]);
				else
					l2.add(a[i]);
			}
			Collections.sort(l1, Collections.reverseOrder());
			Collections.sort(l2, Collections.reverseOrder());
			long pa[] = new long[l1.size() + 1], pb[] = new long[l2.size() + 1];
			for (int i = 1; i <= l1.size(); i++)
				pa[i] = (i == 0 ? 0 : pa[i - 1]) + l1.get(i - 1);
			for (int i = 1; i <= l2.size(); i++)
				pb[i] = (i == 0 ? 0 : pb[i - 1]) + l2.get(i - 1);

			long ans = Integer.MAX_VALUE;
			for (int i = 0; i < pa.length; i++) {
				int idx = findFirst(pb, m - pa[i]);
				if (pa[i] >= m)
					ans = Math.min(ans, i);
				if (idx == -1)
					idx = -Arrays.binarySearch(pb, m - pa[i]) - 1;
				if (idx != pb.length)
					ans = Math.min(ans, 2 * idx + i);
			}
			sb.append(ans == Integer.MAX_VALUE ? -1 : ans);
			sb.append("\n");
		}
		out.print(sb);
		out.close();
	}

	public static int findFirst(long[] a, long target) {
		int idx = -1;
		int s = 0, e = a.length - 1;
		while (s <= e) {
			int m = s + (e - s) / 2;
			if (a[m] >= target)
				e = m - 1;
			else
				s = m + 1;
			if (a[m] == target)
				idx = m;
		}
		return idx;
	}

	static class Scanner {

		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int ni() throws IOException {
			return Integer.parseInt(next());
		}

		public long nl() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nia(int n) throws IOException {
			int a[] = new int[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Integer.parseInt(sa[i]);
			return a;
		}

		public long[] nla(int n) throws IOException {
			long a[] = new long[n];
			String sa[] = br.readLine().split(" ");
			for (int i = 0; i < n; i++)
				a[i] = Long.parseLong(sa[i]);
			return a;
		}

		public int[] sort(int[] a) {
			ArrayList<Integer> l = new ArrayList<>();
			for (int v : a)
				l.add(v);
			Collections.sort(l);
			for (int i = 0; i < a.length; i++)
				a[i] = l.get(i);
			return a;
		}

		public long[] sort(long[] a) {
			ArrayList<Long> l = new ArrayList<>();
			for (long v : a)
				l.add(v);
			Collections.sort(l);
			for (int i = 0; i < a.length; i++)
				a[i] = l.get(i);
			return a;
		}
	}
}