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 int 
#define vi vector<int>
#define vl vector<ll>
#define vvll vector<pair<ll,ll>>
#define fr(i,a,b) for(int i=a;i<b;i++)
#define frr(i,a,b) for(int i=a;i>=b;i--)
#define pb push_back
#define pp pop_back
#define mp make_pair
#define fi first
#define se second
#define lb(a,b) lower_bound(a.begin(),a.end(),b)
#define ub(a,b) upper_bound(a.begin(),a.end(),b)
#define st(a) sort(a.begin(),a.end());
#define s(a) a.size()

using namespace std;

ll gcd(ll a, ll b)
{
    if(a==0)
    return b;
    else if(b==0)
    return a;
    else if(a>b)
    return gcd(b,a%b);
    else
    return gcd(b%a,a);
} 
vector <int> bfs(vector<int> g[], int N) {
vector<int>a;
queue<int>b;
bool c[N]={false};
b.push(0);
c[0]=true;
while(!b.empty())
{ int p=b.front();

a.push_back(p);
for(int i=0;i<g[p].size();i++)
{ if(c[g[p][i]]==false)
   { b.push(g[p][i]);
       c[g[p][i]]=true;}
   }
    
    b.pop();
}
return a;
}
ll dij(ll x , ll y , vvll a[],ll n)
{
    ll dist[n],c,d;
    fr(i,0,n)
    dist[i]=LONG_MAX;
    dist[x]=0;
    bool b[n]={false};
    multiset<pair<ll,ll>>s;
    pair<ll,ll>p;
    s.insert({x,0});
    while(!s.empty())
    { p=*s.begin();
    s.erase(s.begin());
    c=p.first; d=p.second;
    if(b[c]==true)
    continue;
    fr(j,0,a[c].size())
    {
        if(dist[c]+a[c][j].second< dist[a[c][j].first])
        {
            dist[a[c][j].first]=dist[c]+a[c][j].second;
            s.insert({a[c][j].first,dist[a[c][j].first]});
        }
    }
        
    }
    if(dist[y]==LONG_MAX)
    return -1;
    return dist[y];
}
bool cmp(pair<int,int>a,pair<int,int>b)
{
    if(a.first==b.first)
     return (a.second<b.second);
    return (a.first<b.first);
   
}ll nCr(ll n, ll r){
        int dp[r+1][n+1];
        for(int i=0;i<=r;i++)
        {
            for(int j=0;j<=n;j++)
            {
                if(j<i)
                dp[i][j]=0;
                else if(i==0 || i==j)
                dp[i][j]=1;
                else
                dp[i][j]=dp[i-1][j-1]+dp[i][j-1];
              dp[i][j]%=1000000007;
                
            } 
        } return dp[r][n];
    }
int main()
{
    int t;
    cin>>t;
    while(t--)
    { 
       ll n,m,s=0,p=0,sum=0;
       cin>>n>>m;
       long double a[n],b[n];
       vector<pair<long double,int>> c;
       fr(i,0,n)
       {cin>>a[i];sum+=a[i];}
       fr(i,0,n)
       cin>>b[i];
       if(sum<m)
       cout<<-1<<endl;
       else{
       fr(i,0,n)
       c.pb(mp(a[i]/b[i],i));
       st(c);
       int i=n-1,j=n-1;
      
       while(i>=0)
       {
           if((s+a[c[i].second])<m)
           {  p+=b[c[i].second]; s+=a[c[i].second];i--;
               
           }
         
           else
           { p+=b[c[i].second]; s+=a[c[i].second];
               while(s>0)
               {  if((s-a[c[j].second])>=m){
                   s-=a[c[j].second];
                   p-=b[c[j].second];
                   j--;}
                   else
                   break;
               } break;
           }
       }
       cout<<p<<endl;}
       
      
    }
}