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;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        int space[n];
        int points[n];
        for(int i=0;i<n;i++)
        {
            cin>>space[i];
        }
        for(int i=0;i<n;i++)
        {
            cin>>points[i];
        }
        vector< pair<int,int> > v;
        for(int i=0;i<n;i++)
        {
            v.push_back(make_pair(space[i],points[i]));
        }
        sort(v.rbegin(),v.rend());
        vector<int> twoPoint;
        long long sum=0;
        int ind=-1;
        int deduction=0;
        for(int i=0;i<n;i++)
        {
            sum+=v[i].first;
            deduction+=v[i].second;
            if(v[i].second==2)
            {
                twoPoint.push_back(v[i].first);
            }
            if(sum>=m)
            {
                ind=i;
                break;
            }
        }
        if(ind==-1)
        {
            cout<<ind;
        }
        else
        {
            sort(twoPoint.begin(),twoPoint.end());
            vector<int> onePoint;
            for(int i=ind+1;i<n;i++)
            {
                if(v[i].second==1)
                {
                    onePoint.push_back(v[i].first);
                }
            }
            sort(onePoint.begin(),onePoint.end(),greater<int>());
            int i=0;
            int j=0;
            while(i<twoPoint.size() && j<onePoint.size())
            {
                int val=onePoint[j]-twoPoint[i];
                if((sum+val)>=m)
                {
                    sum+=val;
                    i++;
                    j++;
                    deduction--;
                }
                else
                {
                    j++;
                    if(j<onePoint.size())
                    {
                        if((onePoint[j-1]+onePoint[j])>=twoPoint[i])
                        {
                            sum+=(onePoint[j]+onePoint[j-1]-twoPoint[i]);
                            j++;
                            i++;
                        }
                        else
                        {
                            break;
                        }
                    }
                }
            }
            cout<<deduction;
        }
        cout<<endl;
    }
}
