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 pb push_back
#define mp make_pair
#define frall(v) v.begin(),v.end()
#define frrall(v) v.rbegin(),v.rend()
#define fr(i,j,k) for(i=j;i<k;i++)
#define vell vector<ll> 
using namespace std;

int main()
{
    ll t,n,i,j,k,a,b,c,d,p,m,mx,x;
    cin>>t;
    while(t--){
        cin>>n>>m;
        vector<pair<ll,ll>> va(n);
        vell vb(n);
        vell vc;
        vell vd;
        p=0;
        fr(i,0,n) cin>>va[i].first,p+=va[i].first;
        fr(i,0,n) cin>>va[i].second;
        if(p<m) {cout<<-1<<endl; continue;}
        fr(i,0,n) {
            if(va[i].second==1) vc.pb(va[i].first);
            else vd.pb(va[i].first);
        }
        sort(frrall(vc));
        sort(frrall(vd));
        vell dp(2*n+1,-1e13);
        dp[0]=0;
        i=0,j=0,c=0,d=0;
        ll p=vc.size(),q=vd.size();
        ll v=2;
        vell vi(2*n+1,0);
        vell vj(2*n+1,0);
        if(p!=0) dp[1]=vc[i],vi[1]=1;
        //cout<<p<<" "<<q<<" ";
        while(1){
            if(v>=2*n+1) break;
            if(vi[v-1]<p && vj[v-2]<q) {dp[v]=max(dp[v-2]+vd[vj[v-2]],dp[v-1]+vc[vi[v-1]]),x=(dp[v-1]+vc[vi[v-1]]>dp[v-2]+vd[vj[v-2]])?1:0,vi[v]=vi[v-1]+x,vj[v]=vj[v-2]+(1-x);}
            else if(vi[v-1]>=p && vj[v-2]<q)  dp[v]=dp[v-2]+vd[vj[v-2]],vj[v]=vj[v-2]+1,vi[v]=vi[v-1];
            else if(vj[v-2]>=q && vi[v-1]<p) dp[v]=dp[v-1]+vc[vi[v-1]],vi[v]=vi[v-1]+1,vj[v]=vj[v-2];
            else if(vi[v-1]>=p && vj[v-2]>=q) break;
            v++;
        }
        ll f=0;
        //fr(i,1,2*n+1) cout<<dp[i]<<" "<<vi[i]<<" "<<vj[i]<<endl;
        fr(i,1,2*n+1) {if(dp[i]>=m) break;}
        cout<<i<<endl; 
    }
    return 0;
}

