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.
# dp = {}
# inf = [int(1e9+7)]
#
# def solve(list1,num,sum1,index):
#     key = str(num)+"-"+str(sum1)+"-"+str(index)
#     if key in dp.keys():
#         return dp[key]%inf[0]
#     if sum1==0:
#         return 1
#
#     elif index<0:
#         return 0
#     elif num<0:
#         return 0
#
#     else:
#         if list1[index]>sum1:
#             key1 = str(num) + "-" + str(sum1) + "-" + str(index-1)
#             if key1 in dp.keys():
#                 dp[key]  =dp[key1]
#             else:
#                 dp[key1] = solve(list1,num,sum1,index-1)
#                 dp[key] = dp[key1]%inf[0]
#             return dp[key1]%inf[0]
#         else:
#             key1 = str(num-1)+"-"+str(sum1-list1[index])+"-"+str(index-1)
#             key2 = str(num) + "-" + str(sum1) + "-" + str(index-1)
#
#             if key1 not in dp.keys():
#                 dp[key1] = solve(list1,num-1,sum1-list1[index],index-1)
#
#             if key2 not in dp.keys():
#                 dp[key2] = solve(list1,num,sum1,index-1)
#             dp[key] = dp[key2]%inf[0] +dp[key1]%inf[0]
#             return dp[key]%inf[0]
#
#
#
#
#
#
# t = int(input())
# while t!=0:
#
#     n,k = map(int,input().split())
#     list1 = list(map(int,input().split()))
#     list1.sort()
#     dp={}
#     s = sum(list1[len(list1)-k:])
#     ans = solve(list1,k,s,n-1)
#     print(ans)
#
#
#
#
#     t-=1
#
#













t = int(input())
while t!=0:


    n,m = map(int,input().split())
    a = list(map(int,input().split()))
    b = list(map(int,input().split()))



    one=[]
    two=[]
    for i in range(n):
        if b[i]==1:
            one.append(a[i])
        else:
            two.append(a[i])

    one.sort()
    two.sort()

    ans=0
    while len(one)>0 and len(two)>0 and m>0:
        if one[-1]>=m:
            ans+=1
            m-=one[-1]
            one.pop()
            break
        elif m<=two[-1]:
            ans+=2
            m-=two[-1]
            two.pop()
            break
        elif len(one)==1 :
            ans+=2
            m-=two[-1]
            two.pop()
        elif two[-1]>(one[-1]+one[-2]):
            ans+=2
            m-=two[-1]
            two.pop()
        else:
            m-=one[-1]
            ans+=1
            one.pop()
    if m<=0:
        print(ans)
    else:
        while len(one)>0 and m>0:
            ans+=1
            m-=one[-1]
            one.pop()
        if m<=0:
            print(ans)
        else:
            while len(two) > 0 and m > 0:
                ans += 2
                m -= two[-1]
                two.pop()
            if m<=0:
                print(ans)
            else:
                print(-1)









    t-=1
			 	 					 	 	  	 		 	     	