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.
import sys
import bisect as bi
import math
from collections import defaultdict as dd
from types import GeneratorType
##import queue 
##from heapq import heapify, heappush, heappop
##import itertools
##import io
##import os
##import operator
##import random
##sys.setrecursionlimit(10**7)
input=sys.stdin.readline
##input = io.BytesIsoO(os.read(0, os.fstat(0).st_size)).readline
##fo=open("output2.txt","w")
##fi=open("input2.txt","w")
mo=10**9+7
def cin():return map(int,sin().split())
def ain():return list(map(int,sin().split()))
def sin():return input().strip()
def inin():return int(input())

##----------------------------------------------------------------------------------------------------------------------
def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1)
def pref(a,n,f):
    pre=[0]*n
    if(f==0):         ##from beginning
        pre[0]=a[0]
        for i in range(1,n):
            pre[i]=a[i]+pre[i-1]
    else:              ##from end
        pre[-1]=a[-1]
        for i in range(n-2,-1,-1):
            pre[i]=pre[i+1]+a[i]
    return pre
for _ in range(inin()):
    n,m=cin()
    l=ain()
    p=ain()
    one=[]
    two=[]
    for i in range(n):
        if(p[i]==1):
            one+=[l[i]]
        else:
            two+=[l[i]]
    su=sum(l);lo=len(one);lt=len(two);
    if(su<m):
        print(-1)
    elif(su==m):
        print(lo+lt*2)
    else:
        one.sort(reverse=True)
        two.sort(reverse=True)
        ans=n*2
        if(lo):preo=pref(one,lo,0)
        if(lt):pret=pref(two,lt,0)
        if(lo==0):
            pos=bi.bisect(pret,m)
            if(pret[pos-1]!=m):
                    pos+=1
            ans=2*pos
        elif(lt==0):
            pos=bi.bisect(preo,m)
            if(preo[pos-1]!=m):
                    pos+=1
            ans=pos
        else:
            for i in range(lo):
                temp=m-preo[i]
                if(temp<=0):
                    ans=min(ans,i+1)
                    break
                pos=bi.bisect(pret,temp)
                if(pret[pos-1]!=temp):
                    pos+=1
                ans=min(ans,2*pos+i+1)
##            for i in range(lt):
##                temp=m-pret[i]
##                if(temp<=0):
##                    ans=min(ans,2*(i+1))
##                    break
##                pos=bi.bisect(preo,temp)
##                if(preo[pos-1]!=temp):
##                    pos+=1
##                ans=min(ans,2*(i+1)+pos)
##                print(preo[i],temp,pos,ans)
        print(ans)
                
                    
                
     
##-----------------------------------------------------------------------------------------------------------------------#
def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1)
##2 ki power
def pref(a,n,f):
    pre=[0]*n
    if(f==0):         ##from beginning
        pre[0]=a[0]
        for i in range(1,n):
            pre[i]=a[i]+pre[i-1]
    else:              ##from end
        pre[-1]=a[-1]
        for i in range(n-2,-1,-1):
            pre[i]=pre[i+1]+a[i]
    return pre

# in given set of integers
def kadane(A):
    maxSoFar = maxEndingHere = start = end = beg = 0
    for i in range(len(A)):
        maxEndingHere = maxEndingHere + A[i]
        if maxEndingHere < 0:
            maxEndingHere = 0;beg = i + 1
        if maxSoFar < maxEndingHere:
            maxSoFar = maxEndingHere
            start = beg
            end = i
    return (maxSoFar,start,end)  #max subarray sum and its range 
def modFact(n, p):
    if(n<0):return 0
    if n >= p: return 0
    result = 1
    for i in range(1, n + 1):result = (result * i) % p
    return result
def ncr(n, r, p):
    if(n<r or n<0): return 0
    num = den = 1 
    for i in range(r): 
        num = (num * (n - i)) % p ;den = (den * (i + 1)) % p 
    return (num * pow(den,p - 2, p)) % p


##https://github.com/cheran-senthil/PyRival/blob/master/pyrival/misc/bootstrap.py
##write @bootstrap before recursive func
def bootstrap(f, stack=[]):   
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
 
    return wrappedfunc
