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.*;
import static java.lang.Math.*;

public class D {
    static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer tokenizer;
        private InputStream inputStream = System.in;

        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(inputStream));
            tokenizer = null;
        }

        public FastScanner(String inputFilePath) {
            if(inputFilePath != null) {
                try {
                    inputStream = new FileInputStream(new File(inputFilePath));
                } catch (FileNotFoundException e) {
                    throw new RuntimeException(e);
                }
            }
            reader = new BufferedReader(new InputStreamReader(inputStream));
            tokenizer = null;
        }
        public String next() {
            while(tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }

    static class Pair implements Comparable<Pair> {

        public int x, y;

        public Pair() {
        }

        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public int compareTo(Pair o) {
            return o.x - x == 0 ? o.y - y : o.x - x;
        }
    }

//    static String inputFilePath = "src/input.txt";
        static String inputFilePath = null;
    static FastScanner scan = new FastScanner(inputFilePath);

    static int[] arr = new int[200005];
    static Long[] a = new Long[200005];
    static Long[] b = new Long[200005];

    static int n, m, lena, lenb;

    static boolean ok(int val) {
        for (int i = 0; i <= val && i < lena; ++ i) {
            int id = val - i;
            if((id & 1) == 1) continue;
            id /= 2;
            id = min(id, lenb - 1);
            long sum = (long)a[i] + b[id];
            if(sum >= m) return true;
        }
        return false;
    }

    static void sol() {
        n = scan.nextInt();
        m = scan.nextInt();
        a[0] = Long.valueOf(0);
        b[0] = Long.valueOf(0);
        lena = 1;
        lenb = 1;
        int l = 1, r = 0;
        for (int i = 0; i < n; ++ i) arr[i] = scan.nextInt();
        for (int i = 0; i < n; ++ i) {
            int x = scan.nextInt();
            if(x == 1) {
                a[lena] = new Long(arr[i]);
                ++ lena;
            }
            else {
                b[lenb] = new Long(arr[i]);
                ++ lenb;
            }
            r += x;
        }
        Arrays.sort(a, 1, lena, new Comparator<Long>() {
            @Override
            public int compare(Long o1, Long o2) {
                return (int) (o2 - o1);
            }
        });
        Arrays.sort(b, 1, lenb, new Comparator<Long>() {
            @Override
            public int compare(Long o1, Long o2) {
                return (int) (o2 - o1);
            }
        });
//        System.out.println(Arrays.toString(Arrays.copyOfRange(a, 0, lena)));
//        System.out.println(Arrays.toString(Arrays.copyOfRange(b, 0, lenb)));
        for (int i = 1; i < lena; ++ i) a[i] += a[i-1];
        for (int i = 1; i < lenb; ++ i) b[i] += b[i-1];
        int ans = -1;
        while(l <= r) {
           int mid = l + r >> 1;
           if(ok(mid)) {
               if(ans == -1) ans = mid;
               else ans = min(ans, mid);
               r = mid - 1;
           } else l = mid + 1;
        }
        System.out.println(ans);
    }
    public static void main(String[] args) {
        int T = scan.nextInt();
        while(T -- > 0) sol();
    }

}
