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 ld long double
#define fastio ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define pb push_back
#define pf push_front
#define mp make_pair
#define mod 1000000007
#define vll vector<ll>
#define inf 3e18 +7
#define pll pair<ll,ll>
#define ff first 
#define ss second 
#define PI acos(-1)
#define prm 998244353
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds; 
using namespace std;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
/*
 
ll power(ll a, ll b){//a^b
    ll res=1;
    a=a%prm;
    while(b>0){
        if(b&1){res=(res*a)%prm;b--;}
        a=(a*a)%prm;
        b>>=1;
    }
    return res;
}
ll inv(ll y){return power(y,prm-2);}
 
int nCr(int n, int r) 
{ 
   if (r==0) 
      return 1; 
    return (fact[n]* inv(fact[r]) % mod * 
            inv(fact[n-r]) % mod) % mod; 
} 
ll fac[100005];
void init(ll n)
{
	fac[0]=1;
	for(ll i=1;i<=n;i++)fac[i]=(fac[i-1]*i)%prm;
}
*/
/*
ll spf[15000050];
ll cnt[15000050];
void sieve()
{
	for(ll i=2;i*i<=15000030;i++){
		if(spf[i]==0){
			for(ll j=i+i;j<=15000030;j+=i){
				if(spf[j]==0){
					spf[j]=i;
				}
			}
		}
	}
}*/
ll a[200005],b[200005];
int main() 
{		
	fastio;
	ll T=1;
	cin>>T;
	//sieve();	
	while(T--)
	{
		ll n,m;
		cin>>n>>m;
		vll o,t;
		
		ll sum=0;
		for(ll i=0;i<n;i++)
		{
			cin>>a[i];
			sum+=a[i];
		}
		for(ll i=0;i<n;i++)
		{
			cin>>b[i];
			
		}
		if(sum<m)
		{
			cout<<"-1"<<endl;
			continue;
		}
		for(ll i=0;i<n;i++)
		{
			if(b[i]==1)
			{
				o.pb(a[i]);
			}
			else t.pb(a[i]);
		}
		sort(o.begin(),o.end());
		sort(t.begin(),t.end());
		reverse(o.begin(),o.end());
		reverse(t.begin(),t.end());
		ll ans=inf;
		for(ll i=1;i<t.size();i++)
		{
			t[i]+=t[i-1];
		}
		ll res=0;
		
	/*	if(t.size()==0)
		{
			ll ii=lower_bound(o.begin(),o.end(),m-res)-o.begin();
			if(ii<n&&ii>=0)
			{
			ans=min(ans,ii+1);
			cout<<ans<<endl;
			continue;
			}
		}*/
		ll ii=lower_bound(t.begin(),t.end(),m-res)-t.begin();
		if(ii<n&&ii>=0)
		ans=min(ans,2*ii+2);
		for(ll i=0;i<o.size();i++)
		{
			res+=o[i];
			ll id=lower_bound(t.begin(),t.end(),m-res)-t.begin();
			if(res>=m)
			{
				ans=min(ans,i+1);
				break;
			}
			if(id<n&&id>=0)
			ans=min(ans,i+1+2*id+2);
		}
		if(ans!=inf)
		cout<<ans<<endl;
		else
		cout<<"-1"<<endl;
	}
}
