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;

const int N = 200005;
struct Node{
	double a;
	double b;
	bool operator < (const Node & p) const{
		if(a / b != p.a / p.b)
			return (a / b) > (p.a / p.b);
		else{
			return b < p.b;
		}
	}
}node[N];

int main(){
	int t;
	cin >> t;
	
	while(t--){
		int n, k;
		cin >> n >> k;
		
		for(int i=1; i<=n; i++){
			scanf("%lf", &node[i].a);
		}
		
		for(int i=1; i<=n; i++){
			scanf("%lf", &node[i].b);
		}
		
		sort(node + 1, node + 1 + n);
		
		/*
		for(int i=1; i<=n; i++){
			cout << node[i].a << ' ';
		}
		cout << endl;
		for(int i=1; i<=n; i++){
			cout << node[i].b << ' ';
		}
		cout << endl;
		*/
		
		int ans = 0;
		int t = 0;
		int l1 = 0;
		int l2 = 0;
		int i=1;
		int flag = 0;
		for(; i<=n; i++){
			t += node[i].a;
			ans += node[i].b;
			if(node[i].b == 1)
				l1 = node[i].a;
			else{
				l2 = node[i].a;
			}
			if(t >= k){
				if(node[i].b == 2){
					flag = 1;
				}
				break;
			}
		}
		
		if(flag == 1){
		for(;i <=n; i++){
			if(node[i].b == 1 && t - l2 + node[i].a >= k){
				l1 = -1;
				break;
			}
			else if(node[i].b == 1){
				break;
			}
		}
		}
		
		//cout << ans << endl;
		//cout << "l1=" << l1 << ' '<< "t=" << t << endl;
		if(t >= k)
			if(l1 <= t - k && l1 != 0){
				cout << ans - 1 << endl;
			}
			else{
				cout << ans << endl;
			}
		else{
			cout << -1 << endl;
		}
	}
	
	
	return 0;
}