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>
#define int long long
using namespace std;
const int maxn=200005;
int t,n,m,a[maxn],b[maxn],sum1[maxn],sum2[maxn];
int read()//快读 
{
	int x=0,f=1;char ch=getchar();
	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}//读符号 
	while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}//读数字 
	return x*f;
}
signed main(){
	t=read();
	while(t--){
		n=read();m=read();
		for(int i=1;i<=n;++i) a[i]=read();
		for(int i=1;i<=n;++i) b[i]=read();
		int cur1=0,cur2=0;
		for(int i=1;i<=n;++i){
			if(b[i]==1){
				a[++cur1]=a[i];
			}
			else{
				b[++cur2]=a[i];
			}
		}
		sort(a+1,a+1+cur1,greater<int>());
		sort(b+1,b+1+cur2,greater<int>());
		//for(int i=1;i<=cur1;++i) cout<<a[i]<<'|';
		//cout<<"\n";
		//for(int i=1;i<=cur2;++i) cout<<b[i]<<'|';
		//cout<<"\n";
		for(int i=1;i<=cur1;++i) sum1[i]=sum1[i-1]+a[i];
		for(int i=1;i<=cur2;++i) sum2[i]=sum2[i-1]+b[i];
		int mincost=324434523;
		if(sum1[cur1]+sum2[cur2]<m) printf("-1");
		else{
			for(int i=0;i<=cur1;++i){
				if(sum1[i]>=m){
					mincost=min(mincost,i);
					break;
				}
				else{
					int temp=m-sum1[i];
					int t=lower_bound(sum2+1,sum2+1+cur2,temp)-sum2;
					if(t>cur2) continue;
					mincost=min(mincost,i+2*t);
				}
			}
			printf("%d",mincost);
		}
		puts("");
	}
}
