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 ll long long int
#define ve vector<ll>
#define ve_pa vector<pair<ll,ll> >
#define pb push_back
#define mod 1000000007
#define INF 1e18+7
#define ld long double 
#define pll pair<ll,ll>
#define fi first
#define se second
#define mll map<ll,ll>
#define m_ve map<ll,ve>
#define min3(x,y,z) min(x,min(y,z))
#define max3(x,y,z) max(x,max(y,z))
#define W(x) while(x--)
#define ld long double
#define all(x) x.begin(),x.end()
#define popcount(x) __builtin_popcountll(x)
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
//int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
int32_t main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL); 
cout.tie(NULL);
ll t;cin>>t;
while(t--)
{
	ll n,m;cin>>n>>m;
	pll a[n];
	for(int i=0;i<n;i++)
	cin>>a[i].fi;
	for(int i=0;i<n;i++)
	cin>>a[i].se;
	sort(a,a+n);
	ll pre[n],sum[n];
	pre[0]=a[0].fi;
	sum[0]=a[0].se;
	ll ans=INF;
	for(int i=1;i<n;i++)
	{
		pre[i]=pre[i-1]+a[i].fi;
		sum[i]=sum[i-1]+a[i].se;
	}
	for(int i=0;i<n;i++)
	{
		if(pre[i]>=m)
		ans=min(ans,sum[i]);
		ll j=upper_bound(pre,pre+n,pre[i]-m)-pre;
		j--;
		if(j>=0)
		ans=min(ans,sum[i]-sum[j]);
	}
	if(ans==INF)
	ans=-1;
	cout<<ans<<"\n";
	
}
 
 
}