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 <math.h>
 
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    # define M_PI  3.14159265358979323846
 
 
    const int M=1e9+7;
    long long mod(long long x){
        return ((x%M + M)%M);
    }
    long long add(long long a, long long b){
        return mod(mod(a)+mod(b));
    }
    long long mul(long long a, long long b){
        return mod(mod(a)*mod(b));
    }
 
    ll modPow(ll a, ll b){
        if(b==0)
            return 1LL;
        if(b==1)
            return a%M;
        ll res=1;
        while(b){
            if(b%2==1)
                res=mul(res,a);
            a=mul(a,a);
            b=b/2;
        }
        return res;
    }
 
    void solve (){
        ll n,m;
        cin>>n>>m;
        vector<int>a(n);
        vector<int>b(n);
        ll sp=0;
        for(int i=0;i<n;i++) cin>>a[i],sp+=a[i];
        for(int i=0;i<n;i++) cin>>b[i];
        if(sp<m){
            cout<<-1<<"\n";
            return;
        }
        vector<int>c,d;
        for(int i=0;i<n;i++){
            if(b[i]==1) c.push_back(a[i]);
            else d.push_back(a[i]);
        }
        // a has for 1s value and b has for 2s value
        sort(c.rbegin(),c.rend());
        sort(d.rbegin(),d.rend());
        ll ans=1e9;
        int x=c.size();
        int y=d.size();
        vector<ll>ok(y+1);
        for(int i=0;i<y;i++){
            ok[i+1]=ok[i]+d[i];
        }
        ll total=0;
        int pos=lower_bound(ok.begin(),ok.end(),m)-ok.begin();
        if(pos<y+1){
            ans=min(ans,pos*2LL);
        }
        for(int i=0;i<x;i++){
            ll count=i+1;
            total+=c[i];
            ll left=m-total;
            if(left>0){
                pos=lower_bound(ok.begin(),ok.end(),left)-ok.begin();
                if(pos>=y+1) continue;
                count+=(pos*2);
            }
            ans=min(ans,count);
        }
        cout<<ans<<"\n";
    }
    int main(){
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        cout<<fixed;
        cout<<setprecision(10);
//        freopen("timber_input.txt", "r", stdin);
//        freopen("timber_output.txt", "w", stdout);
        int t=1;
        cin>>t;
        for(int i=1;i<=t;i++){
//            cout<<"Case #"<<i<<": ";
            solve();
        }
        return 0;
    }