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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class D {
    final int INF = 0x3f3f3f3f;
    BufferedReader  br;
    PrintWriter     out;
    StringTokenizer st;
    boolean         eof;

    public D() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) {
        new D();
    }

    private void solve() {
        int t = nextInt();
        while (t-- > 0) {
            int n = nextInt();
            int mem = nextInt();
            int[] m = nextIntArr(n);
            int[] v = nextIntArr(n);
            List<Integer> one = new ArrayList<>();
            List<Integer> two = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                if (v[i] == 1) {
                    one.add(m[i]);
                } else {
                    two.add(m[i]);
                }
            }
            one.sort((a, b) -> b - a);
            two.sort((a, b) -> b - a);
            long remove = 0;
            int last = 2;
            int a = 0;
            int b = 0;
            int point = 0;
            while (a < one.size() && b < two.size()) {
                if (remove >= mem) {
                    break;
                }
                if (last == 2) {
                    remove += one.get(a);
                    point += 1;
                    ++a;
                    last = 1;
                    continue;
                }
                if (one.get(a - 1) + one.get(a) < two.get(b)) {
                    --a;
                    remove -= one.get(a);
                    remove += two.get(b);
                    point += 1;
                    ++b;
                    last = 2;
                } else {
                    remove += one.get(a);
                    point += 1;
                    ++a;
                }
            }
            while (b < two.size() && remove < mem) {
                if (a > 0 && remove < mem) {
                    if (remove + two.get(b) - one.get(a - 1) >= mem) {
                        remove += two.get(b) - one.get(a - 1);
                        --a;
                        ++b;
                        point += 1;
                        continue;
                    }
                }
                remove += two.get(b);
                point += 2;
                ++b;
            }
            while (a < one.size() && remove < mem) {
                remove += one.get(a);
                point += 1;
                ++a;
            }
            if (remove < mem) {
                outln(-1);
            } else {
                outln(point);
            }
        }
    }


    private void outln(Object o) {
        out.println(o);
    }

    private void out(Object o) {
        out.print(o);
    }

    public long[] nextLongArr(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextLong();
        }
        return res;
    }

    public int[] nextIntArr(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() {
        return Double.parseDouble(nextToken());
    }


}