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;
 
typedef vector <int> vi;
typedef pair< int ,int > pii;
typedef unordered_set<int, int> hsii; 
typedef unordered_map<int, int> hmii; 

#define endl "\n"
#define deb(x) cout << #x << " " << x << endl;
#define fo(i, n) for(i=0; i<n; i++)
#define Fo(i, k, n) for(i=k; i<n; i++)
#define all(v) v.begin(),v.end()
#define pb push_back
#define mp make_pair
#define FF first
#define SS second
#define INF 1e18
#define ll long long
#define MOD 1000000007
#define clr(val) memset(val,0,sizeof(val))
#define what_is(x) cerr << #x << " is " << x << endl; 
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

template<typename... T>
void read(T&... args) {
    ((cin >> args), ...);
}

template<typename... T>
void write(T&&... args) { //rvalue reference is new to C++ 
    ((cout << args << " "), ...);
}
 
void solve(){
      ll n,m;
      cin>>n>>m;
      vector <pair<ll,ll>> v(n);
      vector <ll> a,b;
      ll ans=INF;
      ll sum=0;
      for(int i=0;i<n;i++)cin>>v[i].FF;
        for(int i=0;i<n;i++){
            cin>>v[i].SS;
            sum+=v[i].FF;
            if(v[i].SS==1){
                a.pb(v[i].FF);
            }
            else b.pb(v[i].FF);
        }
 
        sort(a.begin(),a.end());
        sort(b.begin(),b.end());
        reverse(a.begin(),a.end());
        reverse(b.begin(),b.end());
 
        if(sum<m){
            cout<<-1<<endl;
            return;
        }
 
        sum=0;
        for(int i=0;i<a.size();i++){
            sum+=a[i];
            if(sum>=m){
            ans=min(ans,i+1ll);
        }
        }
        
        sum=0;
        for(int i=0;i<b.size();i++){
            sum+=b[i];
            if(sum>=m){
            ans=min(ans,2ll*(i+1));
        }
        }
        
 
        for(int i=1;i<a.size();i++){
            a[i]+=a[i-1];
        }
        for(int i=1;i<b.size();i++){
            b[i]+=b[i-1];
        }
 
        for(int i=0;i<b.size();i++){
            int x=lower_bound(a.begin(),a.end(),m-b[i])-a.begin();
            if(x<a.size()){
                ans=min(ans,2ll*(i+1) + (x+1));
            }
        }
 
        cout<<ans<<endl;
 
}

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