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.
//package codeforces.contest.c1475;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

/*
https://codeforces.com/problemset/problem/1475/D
 */
public class Task1475_D {

    static void solve() {
        int n = FS.nextInt();
        long m = FS.nextLong();
        long[] a = FS.readArrayL(n);

        List<Long> lessI = new ArrayList<>();
        List<Long> importantI = new ArrayList<>();
        long impSum = 0;
        for (int i = 0; i < n; i++) {
            int b = FS.nextInt();
            if (b == 1) {
                lessI.add(a[i]);
            } else {
                importantI.add(a[i]);
                impSum += a[i];
            }
        }
        lessI.sort(Comparator.naturalOrder());
        importantI.sort(Comparator.naturalOrder());

        Map<Long, Long> map = new TreeMap<>(Comparator.reverseOrder());

        long prevSum = 0;
        long cost = 0;
        long minCost = Long.MAX_VALUE;
        for (int i = lessI.size() - 1; i >= 0; i--) {
            prevSum += lessI.get(i);
            map.put(prevSum, ++cost);
            if (prevSum >= m) {
                minCost = Math.min(minCost, cost);
            }
        }

        for (Map.Entry<Long, Long> it : map.entrySet()) {
            if (it.getKey() + impSum < m) {
                break;
            }

            long sum = it.getKey();
            long newCost = it.getValue();
            for (int i = importantI.size() - 1; i >= 0; i--) {
                sum += importantI.get(i);
                newCost += 2;
                if (sum >= m) {
                    minCost = Math.min(newCost, minCost);
                    break;
                }
            }
        }

        long newCost = 0;
        long sum = 0;
        for (int i = importantI.size() - 1; i >= 0; i--) {
            sum += importantI.get(i);
            newCost += 2;
            if (sum >= m) {
                minCost = Math.min(newCost, minCost);
            }
        }

        if (minCost == Long.MAX_VALUE) {
            FS.pt.println(-1);
            return;
        }
        FS.pt.println(minCost);
    }

    public static void main(String[] args) {
        int T = FS.nextInt();
        for (int tt = 0; tt < T; tt++) {
            solve();
        }
        FS.pt.close();
    }

    static class FS {
        private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        private static StringTokenizer st = new StringTokenizer("");
        static PrintWriter pt = new PrintWriter(System.out);

        static String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        static int nextInt() {
            return Integer.parseInt(next());
        }

        static int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        static double nextDouble() {
            return Double.parseDouble(next());
        }

        static int[][] read2Array(int n, int m) {
            int[][] a = new int[n][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    a[i][j] = nextInt();
                }
            }
            return a;
        }

        static long[][] read2ArrayL(int n, int m) {
            long[][] a = new long[n][m];
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    a[i][j] = nextLong();
                }
            }
            return a;
        }

        void printArr(long[] arr) {
            for (long value : arr) {
                pt.print(value);
                pt.print(" ");
            }
            pt.println();
        }

        static long[] readArrayL(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

        static void printArr(int[] arr) {
            for (int value : arr) {
                pt.print(value);
                pt.print(" ");
            }
            pt.println();
        }

        static void printArrL(int[] arr) {
            for (int value : arr) {
                pt.print(value);
                pt.print(" ");
            }
            pt.println();
        }

        static void print2Arr(int[][] arr, int n, int m) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    pt.print(arr[i][j]);
                    pt.print(" ");
                }
                pt.println();
            }
            pt.println();
        }

        static void print2Arr(long[][] arr, int n, int m) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    pt.print(arr[i][j]);
                    pt.print(" ");
                }
                pt.println();
            }
            pt.println();
        }

        static List<Long> readListLong(int n) {
            List<Long> list = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                list.add(nextLong());
            }
            return list;
        }

        static List<Integer> readListInt(int n) {
            List<Integer> list = new ArrayList<>(n);
            for (int i = 0; i < n; i++) {
                list.add(nextInt());
            }
            return list;
        }

        static <T> void printList(List<T> list) {
            for (T value : list) {
                pt.print(value);
                pt.print(" ");
            }
            pt.println();
        }

        static void close() {
            pt.close();
        }

        static long nextLong() {
            return Long.parseLong(next());
        }
    }

    static class Mat {
        public static long pow(long a, long exp) {
            return (long) Math.pow(a, exp);
        }
    }

    static class NumberTheory {
        public static long gcd(long a, long b) {
            while (b != 0) {
                long tmp = a % b;
                a = b;
                b = tmp;
            }
            return a;
        }

        public static long lcm(long a, long b) {
            return a * b / gcd(a, b);
        }

        public static List<Long> primeFactors(long n) {
            List<Long> ans = new ArrayList<>();
            while (n % 2 == 0) {
                ans.add(2L);
                n /= 2;
            }

            // n must be odd at this point.  So we can
            // skip one element (Note i = i + 2)
            for (long i = 3; i <= Math.sqrt(n); i += 2) {
                // While i divides n, print i and divide n
                while (n % i == 0) {
                    ans.add(i);
                    n /= i;
                }
            }

            // This condition is to handle the case when
            // n is a prime number greater than 2
            if (n > 2) {
                ans.add(n);
            }
            return ans;
        }

        static long fact(long n) {
            if (n == 0 || n == 1) return 1L;
            long r = 1;
            for (int i = 1; i <= n; i++) {
                r *= i;
            }
            return r;
        }
    }

    static class Pair<T, K> {
        T first;
        K second;

        public Pair(T first, K second) {
            this.first = first;
            this.second = second;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair<?, ?> pair = (Pair<?, ?>) o;
            return Objects.equals(first, pair.first) &&
                    Objects.equals(second, pair.second);
        }

        @Override
        public int hashCode() {
            return Objects.hash(first, second);
        }
    }
}
