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 REP(i,n) for(int i=0; i<n;i++)
#define pb push_back
#define ff first
#define ss second
#define ii pair<int,int>
#define vi vector<int>
#define vii vector<ii>
#define lli long long int
using namespace std;

ostream& operator<<(ostream& os, const vector<int> &v){
    for(auto const &i: v){
        os<<i<<" ";
    }
    os<<endl;
    return os;
}
 int main(){
     int cases;cin>>cases;
     while(cases--){
        int n,m;cin>>n>>m;
        vi  ones,twos; int tmp;
        vi apps;vi pesos;
        int unos= 0,doses = 0;
        long long total = 0;
        REP(i,n){
            cin>>tmp;apps.push_back(tmp);
            total+=tmp;
        }
        REP(i,n){
            cin>>tmp;
            if(tmp==1){ones.push_back(apps[i]);unos++;}
            else {twos.push_back(apps[i]); doses++;}
        }
        sort(twos.rbegin(),twos.rend());sort(ones.rbegin(),ones.rend());
        if(total>=m){
                int i=0; int acum = 0; int ans =0;
                while(i<doses && acum<m){
                        acum+=twos[i];i++;ans+=2;
                }
                if (i>0)i--;
                //int ans2=ans;
                int ans2 = 1e9;
                if(acum>=m) ans2 = min(ans,ans2);
              //  cout<<"Total: "<<acum<<" ans "<<ans<<endl;
                REP(j, unos){
                    acum+=ones[j];
                    ans+=1;
                    //cout<<"Total: "<<acum<<" ans "<<ans<<" ans2  "<<ans2<<endl;
                    while(doses>0 && i>=0 && acum-twos[i]>=m ){
                            acum-=twos[i];ans-=2;
                            i--;
                       // cout<<"SubTotal: "<<acum<<" ans "<<ans<<" ans2  "<<ans<<endl;
                    }
                   if(acum>=m) ans2 = min(ans2,ans);
                }
                //cout<<ans<<endl;
                cout<<ans2<<endl;
        }
        else{
            cout<<-1<<endl;
        }
    }
    return 0;
 }





