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 math
for w in range(int(input())):
    n,m=tuple(map(int,input().split()))
    a=list(map(int,input().split()))
    b=list(map(int,input().split()))
    if(sum(a)<m):
        print(-1)
    elif(sum(a)==m):
        print(sum(b))
    else:
        d={}
        for i in range(n):
            if(a[i] not in d):
                d[a[i]]=[b[i]]
            else:
                d[a[i]].append(b[i])
        a=sorted(list(set(a)))
        for i in d:
            d[i]=sorted(d[i])
        x=m 
        i=len(a)-1
        l=[]
        ans=0
        t2=0
        t1=0
        temp=-1
        position=0
        #print(d)
        while(i>=0):
            #print("i",i)
            if(x<=0):
                break
            for j in range(len(d[a[i]])):
                if(x<=0):
                    break
                if(d[a[i]][j]==1 and a[i]>=x):
                    x=-1
                    ans+=1
                    break
                elif(d[a[i]][j]==1 and t2==0):
                    x-=a[i]
                    ans+=1
                elif(d[a[i]][j]==1):
                    if(t1==0):
                        #l.append(d[a[i]])
                        temp=a[i]
                        t1+=1
                    else:
                        if(temp+a[i]>=x):
                            ans+=2
                            x=-1
                            temp=-1
                            break 
                        elif(temp+l[0]>=x ):
                            ans+=3
                            x=-1
                            temp=-1
                            break
                        elif(temp+a[i]>=l[0]):
                            x-=temp+a[i]
                            ans+=2
                            temp=-1
                            t1=0
                        else:
                            x-=l[0]
                            #print("L",l)
                            del(l[0])
                            #print("l",l)
                            ans+=2
                            t2-=1
                            while(t2>0 and x>0):
                                if(temp>=x):
                                    x=-1
                                    ans+=1
                                    break
                                elif(temp+a[i]>=l[0]):
                                    x-=temp+a[i]
                                    ans+=2
                                    temp=-1
                                    t1=0
                                    break
                                else:
                                    x-=l[0]
                                    #print("L",l)
                                    del(l[0])
                                    #print("l",l)
                                    ans+=2
                                    t2-=1
                else:
                    l.append(a[i])
                    t2+=1
                    #print(a[i],len(d[a[i]]))
            i-=1
        if(x>0):
            if(t1==0):
                while(x>0 and t2>0):
                    x-=l[0]
                    del(l[0])
                    ans+=2
                    t2-=1
            else:
                ans1=ans + 0
                ans2=ans + 0
                l1=l.copy()
                l2=l.copy()
                x1=x 
                x2=x
                t21=t2 
                t22=t2
                if(t1==1):
                   only_left_one = temp
                   x1-=temp 
                   ans1+=1
                #print("YO")
                while(x1>0 and t21>0):
                   x1-=l1[0]
                   del(l1[0])
                   ans1+=2
                   t21-=1
                while(x2>0 and t22>0):
                    x2-=l2[0]
                    ans2+=2 
                    del(l2[0])
                    t22-=1
                if(x2>0):
                    x2-=temp 
                    ans2+=1
                ans=min(ans1,ans2)
        print(ans)
        #print(l,a)
        #print(d)