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.
# Problem 1475D - Cleaning The Phone

def solve():
    inp = input().rstrip().split(" ")
    n, m = int(inp[0]), int(inp[1])
    a_arr = input().rstrip().split(" ")
    b_arr = input().rstrip().split(" ")
    
    sum_reg = 0
    sum_imp = 0
    reg_apps = []
    imp_apps = []
    for i in range(n):
        if b_arr[i] == "1":
            reg_apps.append(int(a_arr[i]))
            sum_reg += int(a_arr[i])
        else:
            imp_apps.append(int(a_arr[i]))
            sum_imp += int(a_arr[i])
    
    reg_apps.sort()
    imp_apps.sort()
    
    if sum_reg + sum_imp < m:
        print(-1)
        return
    
    tot_removed = 0
    min_points_lost = 0
    
    while sum_reg < m - tot_removed:
        tot_removed += imp_apps.pop()
        min_points_lost += 2
    
    while tot_removed < m and len(reg_apps)+len(imp_apps) > 0:
        if len(reg_apps) == 0:
            tot_removed += imp_apps.pop()
            min_points_lost += 2
        elif len(imp_apps) == 0:
            tot_removed += reg_apps.pop()
            min_points_lost += 1
        else:
            if reg_apps[-1] >= imp_apps[-1]:
                tot_removed += reg_apps.pop()
                min_points_lost += 1
            else:
                if tot_removed + reg_apps[-1] >= m:
                    tot_removed += reg_apps.pop()
                    min_points_lost += 1
                elif len(reg_apps) > 1 and reg_apps[-1] + reg_apps[-2] > imp_apps[-1]:
                    tot_removed += reg_apps.pop()
                    tot_removed += reg_apps.pop()
                    min_points_lost += 2
                else:
                    tot_removed += imp_apps.pop()
                    min_points_lost += 2
    
    print(min_points_lost)

def main():
    T = int(input())
    for c in range(T):
        solve()

main()
