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.
"""
ID: happyn61
LANG: PYTHON3
PROB: loan
"""

#from collections import defaultdict
import sys
import heapq
from collections import deque

def primes(n):
    """ Returns  a list of primes < n """
    sieve = [True] * n
    for i in range(3,int(n**0.5)+1,2):
        if sieve[i]:
            sieve[i*i::2*i]=[False]*((n-i*i-1)//(2*i)+1)
    return [2] + [i for i in range(3,n,2) if sieve[i]]
def factor(n,l,d):
    F=False
    for p in l:
        if n%p==0:
            if p in d:
                d[p]+=1
            else:
                d[p]=1
            F=True
            break
    if F:
        factor(n//p,l,d)
            
#fin = open ('loan.in', 'r')
#fout = open ('loan.out', 'w')
#print(dic["4734"])
def find(parent,i):


    if parent[i] != i: 
        parent[i]=find(parent,parent[i]) 
    return parent[i] 

        # A utility function to do union of two subsets 
def union(parent,rank,xx,yy): 
    x=find(parent,xx)
    y=find(parent,yy)
    if rank[x]>rank[y]:
        parent[y]=x
    elif rank[y]>rank[x]:
        parent[x]=y
    else:
        parent[y]=x
        rank[x]+=1
ans=0

#NQ=sys.stdin.readline().strip().split()

n=int(sys.stdin.readline().strip())
#N1=int(NQ[0])
#N2=int(NQ[1])
#N3=int(NQ[1])

for j in range(n):
    AB=sys.stdin.readline().strip().split()
    #a=int(sys.stdin.readline().strip())
    a=int(AB[0])
    b=int(AB[1])
    #k=int(AB[2])
    al=[0 for i in range(a)]
    bl=[0 for i in range(b)]
    ak=sys.stdin.readline().strip().split()
    bk=sys.stdin.readline().strip().split()
    h1=[]
    h2=[]
    ans=0
    for i in range(a):
        ak[i]=int(ak[i])
        
        bk[i]=int(bk[i])
        if bk[i]==1:
            h1.append(ak[i])
        else:
            h2.append(ak[i])
            
    if (sum(h1)+sum(h2))<b:
        print(-1)
        continue
    h1.sort()
    h2.sort()

    while b>0:
        sa=0
        sb=0
        if len(h1)>2:
            sa=h1[-1]+h1[-2]
            if len(h2)>0:
                sb=h2[-1]
            if sb>=sa:
                b-=h2.pop()
            else:
                b-=h1.pop()
                b-=h1.pop()
            ans+=2
        else:
            if h1 and h1[-1]>=b or not h2:
                ans+=1
                b-=h1.pop()
            elif len(h1)==2 and sum(h1)>=b:
                ans+=2
                b-=h1.pop()
                b-=h1.pop()
            else:
                ans+=2
                
                b-=h2.pop()
            #print(b,h1,h2)
    print(ans)


#    print("".join(l))
    #print(s)
    #l=sys.stdin.readline().strip().split()
    #s1=sys.stdin.readline().strip()
    #s2=sys.stdin.readline().strip()
    
        #print(pre,post,ll,rr,m1,m2,pre[ll],post[n-1-rr],post[n-1-rr][2])
    #if F:
    #    print("yes")
    #else:
    #    print("no")
    #return True
    
#for x,y in occupy:
#    l[x][y]="X"
#for ll in l:
#    print("".join(ll))

