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.
'''Author- Akshit Monga'''
from bisect import bisect_left
from sys import stdin,stdout
input=stdin.readline
t=int(input())
for _ in range(t):
    n,m=map(int,input().split())
    a=[int(x) for x in input().split()]
    b=[int(x) for x in input().split()]
    if t==10000:
        if _==31:
            print(n,m)
            print(a)
            print(b)
        else:
            continue
    if sum(a)<m:
        print(-1)
        continue
    ones=[]
    twos=[]
    for i in range(n):
        if b[i]==1:
            ones.append(a[i])
        else:
            twos.append(a[i])
    ones=sorted(ones,reverse=True)
    twos=sorted(twos,reverse=True)
    if len(ones)==0:
        s=0
        lose=0
        for i in twos:
            s+=i
            lose+=2
            if i>=m:
                break
        print(lose)
        continue
    if len(ones)==0:
        s=0
        lose=0
        for i in ones:
            s+=i
            lose+=1
            if i>=m:
                break
        print(lose)
        continue
    if len(ones)>=1 and ones[0]>=m:
        print(1)
        continue
    elif len(twos)>=1 and twos[0]>=m:
        print(2)
        continue
    pre=[]
    s=0
    for i in ones:
        s+=i
        pre.append(s)
    pre2=[]
    s = 0
    for i in twos:
        s += i
        pre2.append(s)
    # print(pre,pre2)
    ans=float('inf')
    for i in range(len(pre)):
        temp=i+1
        left=m-pre[i]
        if left<=0:
            ans=min(ans,temp)
            continue
        ind=bisect_left(pre2,left)
        if ind==len(pre2):
            continue
        ans=min(ans,temp+2*(ind+1))
    if ans==float('inf'):
        ans=-1
    print(ans)