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.sql.Array;
import java.util.*;

public class Main {


    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        int tc = sc.nextInt();
        main:
        while (tc-- > 0) {
            int n = sc.nextInt(), m = sc.nextInt();
            int[] a = sc.nextIntArray(n);
            ArrayList<Integer> o = new ArrayList<>(), t = new ArrayList<>();
            for (int i = 0; i < n; i++)
                if (sc.nextInt() == 1) o.add(i);
                else t.add(i);
            Collections.sort(o, (x, y) -> a[y] - a[x]);
            Collections.sort(t, (x, y) -> a[y] - a[x]);
            long[] one = new long[o.size()], two = new long[t.size()];
            int ans = -1;
            for (int i = 0; i < o.size(); i++) {
                one[i] = a[o.get(i)];
                if (i != 0) one[i] += one[i - 1];
                if (one[i] >= m) {
                    if (ans == -1) ans = i + 1;
                    else ans = Math.min(ans, i + 1);
                }
            }
            for (int i = 0; i < t.size(); i++) {
                two[i] = a[t.get(i)];
                if (i != 0) two[i] += two[i - 1];
                if (two[i] >= m) {
                    if (ans == -1) ans = (i + 1) * 2;
                    else ans = Math.min(ans, (i + 1) * 2);
                }
            }
            for (int i = 0; i < two.length && two[i] < m; i++) {
                long need = m - two[i];
                int lo = 0, hi = one.length - 1;
                while (lo <= hi) {
                    int mid = lo + hi >> 1;
                    if (one[mid] >= need) {
                        int cand = (i + 1) * 2 + (mid + 1);
                        if (ans == -1) ans = cand;
                        else ans = Math.min(ans, cand);
                        hi = mid - 1;
                    } else lo = mid + 1;
                }
            }
            out.println(ans);
        }
        out.flush();
        out.close();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

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

        public int[] nextIntArray(int n) throws IOException {
            int[] ans = new int[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        public long[] nextLongArray(int n) throws IOException {
            long[] ans = new long[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextLong();
            return ans;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] ans = new Integer[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }


    }
}