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.
#TESTING..
in_length = int(input())
final_ouput = []

for i in range(in_length):
  [num_apps, mem_needed] = input().split()
  mem_used = [int(el) for el in input().split()]
  conv_points = input().split()
  
  total_amt = 0;
  
  
  low_apps = []
  high_apps = []
  
  for j in range(int(num_apps)):
    if conv_points[j] == '1':
      low_apps.append(mem_used[j])
      total_amt += mem_used[j]
    else:
      high_apps.append(mem_used[j])
      total_amt += mem_used[j]
  
  if total_amt < int(mem_needed):
    final_ouput.append(-1)
    continue
      
  s_low_apps = sorted(low_apps)
  low_apps = s_low_apps[:]
  s_high_apps = sorted(high_apps)
  
  efficient = []
  low_sum = 1
  high_sum = 1
  
  for j in range(int(num_apps)):
    if low_sum == 0 and high_sum == 0:
      break
    if len(s_high_apps) > 0:
      high_sum = s_high_apps[-1]
    else:
      high_sum = 0
      
    if len(s_low_apps) > 1:
      low_sum = s_low_apps[-1] + s_low_apps[-2]
      if low_sum > high_sum:
        efficient.append(s_low_apps.pop())
        efficient.append(s_low_apps.pop())
      else:
        efficient.append(s_high_apps.pop())  
    
    elif len(s_low_apps) == 1:
      low_sum = s_low_apps[-1]
      if low_sum >= high_sum:
        efficient.append(s_low_apps.pop())
      else:
        efficient.append(s_high_apps.pop())
    else:
      low_sum = 0
      if high_sum != 0:
        efficient.append(s_high_apps.pop())
        
      
       
  
  curr_needed = int(mem_needed)
  points_lost = 0;
  
  for j in range(int(num_apps)):
    if curr_needed <= 0:
      break
    
    if len(low_apps) > 0:
      best_low = low_apps[-1]
    else:
      best_low = 0
      
    if best_low >= curr_needed:
      curr_needed -= best_low
      points_lost += 1
      break
    elif len(efficient) > 1 and efficient[1] >= curr_needed:
      curr_needed -= efficient[1]
      points_lost += 2
      break
      
    else:
      if efficient[0] == best_low:
        curr_needed -= low_apps[-1]
        low_apps.pop()
        efficient.pop(0)
        points_lost += 1
      else:
        curr_needed -= efficient[0]
        points_lost += 2
        efficient.pop(0)
      
  final_ouput.append(points_lost)
   

for out in final_ouput:
  print(out)

#TESTING..