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 ll long long
#define ull unsigned long long
#define me(a,x) memset(a,x,sizeof(a))
#define sf scanf
#define pf printf
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define per(i,x,y) for(int i=x;i>=y;i--)
#define gcd __gcd
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const ll llINF=9223372036854775807;
const int maxn=2e5+10;
const double pi=acos(-1.0);
using namespace std;
ll n,m;
ll a[maxn],b[maxn],num[maxn];

bool cmp(ll x,ll y){
	return x>y;
} 

void solve()
{
	int x;
	cin>>n>>m;
	ll sum=0;
	rep(i,1,n){
		cin>>num[i];
		sum+=num[i];
	}
	int cnt1=1,cnt2=1;
	rep(i,1,n){
		cin>>x;
		if(x==1){
			a[cnt1++]=num[i];
		}else{
			b[cnt2++]=num[i];
		}
	}
	if(sum<m){
		cout<<-1<<endl;
		return ;
	}
	sort(a+1,a+cnt1,cmp);
	sort(b+1,b+cnt2,cmp);
	a[0]=b[0]=a[cnt1]=b[cnt2]=0;
	rep(i,1,cnt1-1){
		a[i]+=a[i-1];
	}
	rep(i,1,cnt2-1){
		b[i]+=b[i-1];
	}
	int res=INF;
	rep(i,0,cnt1-1){
		ll tmp=m-a[i];
		int id=lower_bound(b+1,b+cnt2,tmp)-b;
		if(a[i]+b[id]>=m)
		res=min(res,i+id*2);
		id--;
		if(a[i]+b[id]>=m){
			res=min(res,i+id*2);
		}
//		cout<<"!"<<i<<" "<<id<<" "<<res<<endl;
	}
	cout<<res<<endl;
}

int main()
{
	IOS;
	int t;
	cin>>t;
	while(t--){
		solve();
	}
	return 0;
}