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.
#include<bits/stdc++.h>
using namespace std;

typedef long long int in;

int main() 
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	in t;
	cin>>t;

	while(t--)
	{
		in n,m;
		cin>>n>>m;

		in a[n],b[n],c[n],d[n],q=0,r=0,tot=0,i,ans=1e18;

		for(i=0;i<n;++i)
		cin>>a[i];

		for(i=0;i<n;++i)
		cin>>b[i];

		for(i=0;i<n;++i)
		if(b[i]==1)
		c[q++]=a[i];
		else
		d[r++]=a[i];

		sort(c,c+q,greater<in>());
		sort(d,d+r,greater<in>());

		for(i=1;i<r;++i)
		d[i]+=d[i-1];

		for(i=0;i<q;++i)
		{
			tot+=c[i];

			in point=1e18;

			if(tot>=m)
			{
				point=i+1;
			}
			else
			{
				in l=0,h=r,mid;

				while(l<h)
				{
					mid=(l+h)/2;

					if(d[mid]>=m-tot)
					h=mid;
					else
					l=mid+1;
				}

				if(h==0&&d[h]<m-tot)
				++h;

				if(h<r)
				point=i+1+2*(h+1);
			}

			ans=min(ans,point);
		}

		tot=0;

		in l=0,h=r,mid,point=1e18;

		while(l<h)
		{
			mid=(l+h)/2;

			if(d[mid]>=m-tot)
			h=mid;
			else
			l=mid+1;
		}

		if(h==0&&d[h]<m-tot)
		++h;

		if(h<r)
		{
			point=2*(h+1);
     		ans=min(ans,point);
		}

		if(ans==1e18)
		ans=-1;

		cout<<ans<<"\n";
	}

	return 0;
}