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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define int           long long
#define start         ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define pb            push_back
#define fr            first
#define sc            second     
#define rep(i,a,b)    for(int  i=a;i<b;i++)
#define repr(i,a,b)    for(int  i=b-1;i>=a;i--)
#define M             1000000007
#define mod           998244353
#define PI 3.14159265
using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> newset;

//int dx[4]={0,0,1,-1};
//int dy[4]={1,-1,0,0};


int power(int x, int y, int m) 
{ 
    if (y == 0) 
        return 1; 
    int p = power(x, y/2, m) % m; 
    p = (p * p) % m; 
    return (y%2 == 0)? p : (x * p) % m; 
} 

int modInverse(int a,int m) {
    return power(a,m-2,m);
}

void solve()
{
	int n,m;
	cin>>n>>m;
	int a[n],c[n];
	vector<pair<int,int>>v;
	rep(i,0,n)
	cin>>a[i];
	rep(i,0,n)
	{
		cin>>c[i];
		v.pb({a[i],c[i]});
	}
	sort(v.begin(),v.end());
	reverse(v.begin(),v.end());
	int p=0,q,ans=0;
	rep(i,0,n)
	{
		p+=v[i].fr;
		ans+=v[i].sc;
		if(p>=m)
		{
			q=i;
			break;
		}
	}
	if(p<m)
	{
		cout<<"-1"<<"\n";
		return;
	}
	vector<int>one,two;
	rep(i,0,q+1)
	{
		if(v[i].sc==2)
		{
			two.pb(v[i].fr);
		}
	}
	rep(i,q+1,n)
	{
		if(v[i].sc==1)
		{
			one.pb(v[i].fr);
		}
	}
	if(one.size()==0 || two.size()==0)
	{
		cout<<ans<<"\n";
		return;
	}
	int i=one.size()-1,j=two.size()-1,e=p-m;
	while(j>=0 && i>=0)
	{
		int d=two[j]-e;
		while(i>=0 && one[i]<d)
			i--;
		if(i<0)
			break;
		ans-=1;
		p-=two[j];
		p+=one[i];
		e=p-m;
		j--;
		i--;
	}
	cout<<ans<<"\n";
}  
    


int32_t main()
{
    start
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int t=1;
    cin>>t;
    while(t--)
    {
        solve();
    }
}