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.

def main():
    
    t=int(input())
    allAns=[]
    for _ in range(t):
        n,m=readIntArr()
        oneConvenientPt=[] #stores memory for apps with 1 conveient pt
        twoConvenientPts=[]
        
        a=readIntArr()
        b=readIntArr()
        for i in range(n):
            if b[i]==1:
                oneConvenientPt.append(a[i])
            else:
                twoConvenientPts.append(a[i])
        oneConvenientPt.sort(reverse=True)
        twoConvenientPts.sort(reverse=True)
        
        twoPrefSums=twoConvenientPts.copy()
        for i in range(1,len(twoPrefSums)):
            twoPrefSums[i]+=twoPrefSums[i-1]
        
        #try 0 1pt app, 1 1pt app, etc... start with the app with most memory
        minConvenientPts=inf
        oneTotal=0
        oneCnts=0
        oneConvenientPt=[0]+oneConvenientPt #to test the case where no elements are taken from a
#        print(oneConvenientPt)####
#        print(twoConvenientPts)######
#        print(twoPrefSums)
        for x in oneConvenientPt:
            oneTotal+=x
            
            if oneTotal>=m: #don't need any b
                minConvenientPts=min(minConvenientPts,oneCnts)
            else:
            
                #binary search to find bElements
                b=n
                i=-1 #i is two's index but also the 1 less than the number of two elements to take
                while b>0:
                    while i+b<len(twoPrefSums) and twoPrefSums[i+b]+oneTotal<m:
                        i+=b
                    b//=2
                i+=1
                if i<len(twoPrefSums):
                    twoCnts=i+1
                    convenientPts=oneCnts+2*twoCnts
#                    print('oneCnts:{} twoCnts:{} convenientPts:{} atotal:{}'.format(oneCnts,twoCnts,convenientPts,oneTotal))#########
                    minConvenientPts=min(minConvenientPts,convenientPts)
            oneCnts+=1
            
        if minConvenientPts==inf:
            allAns.append(-1)
        else:
            allAns.append(minConvenientPts)
    multiLineArrayPrint(allAns)
    
    return
    
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
#import sys
#input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
 
def oneLineArrayPrint(arr):
    print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
    print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
    print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
 
def readIntArr():
    return [int(x) for x in input().split()]
 
inf=float('inf')
MOD=10**9+7
 
main()