1475_D. Cleaning the Phone
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.
```python
#lösningsmängd är nedåtbegränsad och ordnad. -> optimal minsta existerar i kontext.
#Vet att det är sant att lösning består av x stna 1-cost x tillhör [0..all(1-cost)]
#för x stna 1-cost bestäms y stna 2-cost entydligt.
#itererar alla x, försök i varje steg reducera y från mx(2-cost)

#same hold tru if conv points 1 and 3?
#4 number sum in list eq. x?

#scaleas upp till 3? hör med markus
#n2 , likt hitta tre tal vars summa är...
#mot knapsack beroende på m
for _ in range(int(input())):
    n,m = map(int,input().split())
    memoryCost = list(map(int,input().split()))
    convPoints = list(map(int,input().split()))

    cost1 = list()
    cost2 = list()

    for mem,conv in zip(memoryCost,convPoints):
        if conv == 1:
            cost1.append(mem)
        else:
            cost2.append(mem)

    cost1.sort(reverse=True)
    cost2.sort(reverse=True)

    #vi ska ha x stna 1-cost. Då följer y-stna 2-cost. (reducerat)
    #börja med alla 2-cost och iterera alla 0-n stna 1-cost
    memory = sum(cost2)
    convP = 2*(len(cost2))
    twoCostPointer = (len(cost2))-1
    ans = float("inf")
    #få till 2 extra iter
    for x in range(-1,len(cost1)):
        #add x, remove maximalt med 2cost
        if x >= 0:
            convP += 1
            memory += cost1[x]
        
        while(twoCostPointer >= 0 and memory - cost2[twoCostPointer] >= m):
            memory -= cost2[twoCostPointer]
            twoCostPointer -= 1
            convP -= 2
        #if
        if memory >= m:
            ans = min(ans,convP)

    print(ans if ans != float("inf") else -1)   

'''
1
5 7
5 3 2 1 4
2 1 1 2 1
'''
```