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;
#define ll long long
#define ull unsigned long long
#define ld long double
#define mod 1e9 + 7
#define INF 1000000000000000003
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    ll ct=0;
    for(int k=1;k<=T;k++)
	{
		// your code
		ll n,m;
		cin >> n >> m;
		vector<ll> a(n);
		vector<int> b(n);
		for(int i=0;i<n;i++)
			cin >> a[i];
		for(int i=0;i<n;i++)
			cin >> b[i];
		
		vector<ll> a1,a2;
		a1.reserve(n);
		a2.reserve(n);
		ll sum=0;
		for(int i=0;i<n;i++)
		{
			if(b[i]==1)
				a1.push_back(a[i]);
			else
				a2.push_back(a[i]);
			sum+=a[i];
		}
		sort(a1.begin(),a1.end(),greater<>());
		sort(a2.begin(),a2.end(),greater<>());
		
		if(sum<m)
			cout << "-1\n";
		else{
			int cost=0;
			sum=0;
			int a1s=a1.size(),a2s=a2.size();
			int i=0,j=0;
			while(sum<m)
			{
				if(i==a1s)
				{
					sum+=a2[j];
					cost+=2;
					j++;
					continue;
				}
				if(j==a2s)
				{
					sum+=a1[i];
					cost+=1;
					i++;
					continue;
				}
				if(i==(a1s-1))
				{
					if((sum+a1[i])>=m)
					{
						sum+=a1[i];
						cost++;
						i++;
					}
					else if((sum+a1[i]+a2[j])>=m && (sum+a2[j])<m)
					{
						sum+=a1[i]+a2[j];
						cost+=3;
						i++;
						j++;
					}
					else{
						sum+=a2[j];
						cost+=2;
						j++;
					}
					continue;
				}
				if((sum+a1[i])>=m)
				{
					sum+=a1[i];
					cost++;
					i++;
					continue;
				}
				if((sum+a2[j])>=m)
				{
					sum+=a2[j];
					cost+=2;
					j++;
					continue;
				}
				if((sum+a1[i]+a1[i+1])>=m)
				{
					sum+=a1[i]+a1[i+1];
					cost+=2;
					i+=2;
					continue;
				}
				if((sum+a1[i]+a2[j])>=m)
				{
					sum+=a1[i]+a2[j];
					cost+=3;
					i++;
					j++;
					continue;
				}
				if(a1[i]>=a2[j])
				{
					sum+=a1[i];
					cost++;
					i++;
					continue;
				}
				int temp1=a1[i]+a1[i+1];
				int temp2=a2[j];
				if(temp1>temp2)
				{
					sum+=temp1;
					cost+=2;
					i+=2;
				}
				else{
					sum+=temp2;
					cost+=2;
					j++;
				}
			}
			ct++;
			if(T==5)
				cout << cost << "\n";
			else if(k==258){
				cout << m << "\n";
				for(int i=0;i<n;i++)
					cout << a[i] << " ";
				cout << "\n";
				for(int i=0;i<n;i++)
					cout << b[i] << " ";
				cout << "\n" << cost << "\n";
			}
		}
		
    }
 	return 0;
}
