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>
// no of bits to store no 2e5 is 19
// 2e5 ka 5 times square root lene se 2 or 3 ke pass pahunch jate hai
using namespace std;

#define int               long long
#define mod               1000000007
#define MOD               1000000007
#define MAX               1000000000
#define ff                first
#define ss                second
#define mp                make_pair
#define pb                push_back
#define pii               pair<int,int>
#define sp(x,y)           fixed<<setprecision(y)<<x
#define setbits(x)        __builtin_popcount(x)
#define fast_in_out       ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define ll                long long

//   powermod(a,b,m) ==  powermod(a,b%(m-1),m)  when m is prime fermat's theorem a^(p-1) %p=1 p is prime
int powermod(int a,int b,int c) //  gives (a^b)%c
{
    int ans = 1,tempans=a%c;
    while(b>0)
    {
        if(b&1==1)
        {
            ans=(ans*tempans)%c;
        }
             tempans=(tempans*tempans)%c;
             b=b>>1;
    }
    return ans;
}

int modinverse(int a,int m)   // gives a inverse wrt m and m is prime
{
    return powermod(a,m-2,m);
}

ll add(ll x, ll y) {
   x += y;
   if (x >= mod) return x - mod;
   return x;
}
ll sub(ll x, ll y) {
   x -= y;
   if (x < 0) return x + mod;
   return x;
}
ll mult(ll x, ll y) {
   return (x * y) % mod;
}
// my template                                              m
// prime factorization                                      pf
// ncr and factorials                                       ncr
// matrix exponentiation of fibonacci series in log(n)      matexpofib
// cycles in a graph                                        cycle
// prime factorization in log(n)                            spf



void solve()
{
   int n,m;
   cin>>n>>m;
   int mem[n],b[n];
   for(int i=0;i<n;i++)
   {
      cin>>mem[i];
   }
   for(int i=0;i<n;i++)
   {
      cin>>b[i];
   }
   vector<int> one,two;
   for(int i=0;i<n;i++)
   {
      if(b[i]==1)
      {
         one.pb(mem[i]);
      }
      else
      {
         two.pb(mem[i]);
      }
   }
   sort(one.begin(),one.end(),greater<int>());
    sort(two.begin(),two.end(),greater<int>());
    one.insert(one.begin(),0);
    two.insert(two.begin(),0);
    for(int i=1;i<one.size();i++)
    {
       one[i]+=one[i-1];
    }
    for(int i=1;i<two.size();i++)
    {
       two[i]+=two[i-1];
    }
    int ans=1e15;
    for(int i=0;i<one.size();i++)
    {
       int temp=one[i];
       int rem=m-temp;
       int x=lower_bound(two.begin(),two.end(),rem)-two.begin();
     //  cout<<temp<<" "<<rem<<" "<<x<<endl;
       if(x!=(int)two.size())
       {
          ans=min(ans,i+2*x);
       }
    }
    if(ans==1e15)
    {
       cout<<-1<<endl;
    }
    else
    {
       cout<<ans<<endl;
    }





}


int32_t main()
{
  fast_in_out;

  int t=1;cin>>t;
  int count=0;


   while(t--)
   {
       count++;
       solve();
   }
}
