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.*;

public class D {
    private static FastReader fr = new FastReader();
    private static PrintWriter out=new PrintWriter(System.out);
    private static Random random = new Random();

    public static void main(String[] args) throws IOException {
        StringBuilder sb = new StringBuilder();
        // code goes here
        int t = fr.nextInt();
        while (t-- > 0){
            int n = fr.nextInt();
            long m = fr.nextInt();
            int[] a = fr.nextIntArray(n);
            int[] b = fr.nextIntArray(n);
            List<Integer> ones = new ArrayList<>();
            List<Integer> twos = new ArrayList<>();
            for(int i = 0; i < n; i++){
                if(b[i] == 1){
                    ones.add(a[i]);
                } else {
                    twos.add(a[i]);
                }
            }

            ones.sort(null);

            twos.sort(null);

            long sum = 0;
            long ans = Long.MAX_VALUE;
            List<Long> ll = new ArrayList<>();
            long temp = 0;
            for(int i = twos.size() - 1; i >= 0; i--){
                temp += twos.get(i);
                ll.add(temp);
            }
            for(int i = ones.size() - 1; i >= 0; i--){
                sum += ones.get(i);
                long req = m - sum;
                if(req <= 0){
                    ans = Math.min(ans, ones.size() - i);
                } else {
                    int lo = 0, hi = ll.size() - 1;
                    while (lo <=  hi){
                        int mid = lo + (hi - lo)/2;
                        long x = ll.get(mid);
                        if(x >= req){
                            ans = Math.min(ans, ones.size() - i + (2 * (mid + 1)));
                            hi = mid - 1;
                        } else {
                            lo = mid + 1;
                        }
                    }
                }
            }

            if(ones.size() == 0){
                long req = m;
                int lo = 0, hi = ll.size() - 1;
                while (lo <=  hi){
                    int mid = lo + (hi - lo)/2;
                    long x = ll.get(mid);
                    if(x >= req){
                        ans = Math.min(ans,(2 * (mid + 1)));
                        hi = mid - 1;
                    } else {
                        lo = mid + 1;
                    }
                }
            }

            if(ans != Long.MAX_VALUE){
                sb.append(ans).append("\n");
            } else {
                sb.append("-1\n");
            }
        }
        System.out.print(sb.toString());
    }

    static void ruffleSort(int[] a) {
        int n=a.length;//shuffle, then sort
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n), temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        Arrays.sort(a);
    }

    static class FastReader{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");

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

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

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

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

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

    static class Pair<A, B>{
        A first;
        B second;
        public Pair(A first, B second){
            this.first = first;
            this.second = second;
        }
    }

    static long mod(String num, long a)
    {
        // Initialize result
        long res = 0;

        // One by one process all digits of 'num'
        for (int i = 0; i < num.length(); i++)
            res = (res*10 +  num.charAt(i) - '0') %a;

        return res;
    }

    static long binomialCoeff(long n, long k, long MOD)
    {
        long res = 1;

        // Since C(n, k) = C(n, n-k)
        if (k > n - k)
            k = n - k;

        // Calculate value of
        // [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
        for (int i = 0; i < k; ++i) {
            res *= (n - i);
            res /= (i + 1);
            res %= MOD;
        }

        return res;
    }

    static long power(long x, long y, long p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, long p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static long nCrModPFermat(int n, int r,
                              long p)
    {

        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        long[] fac = new long[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;

        return (fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
                % p;
    }
}
