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.
		//Iamskk//
#include <iostream>
#include <iosfwd>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cassert>
#include <cctype>
#include <climits>
#include <vector>
#include <bitset>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <deque>
#include <string>
#include <list>
#include <iterator>
#include <sstream>
#include <complex>
#include <fstream>
#include <functional>
#include <numeric>
#include <utility>
#include <algorithm>
#include <assert.h>
#include <unordered_map>
using namespace std;
#define IOS ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(0);
#define endl '\n'
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define rall(c) (c).rbegin(),(c).rend()
#define ai(arr,n) for(int i=0;i<n;i++)cin>>arr[i];
#define ao(arr) for(auto asdfasdfasdf:arr) cout<<asdfasdfasdf<<" ";
#define mi(arr,m,n) for(int i=0;i<m;i++){ for(int j=0;j<n;j++) cin>>arr[i][j];}
#define mo(arr,m,n) for(int i=0;i<m;i++){ for(int j=0;j<n;j++) cout<<arr[i][j]<<" "; cout<<endl;}
#define countsetbits(x) __builtin_popcount(x)
#define ll long long
#define debug cout<<"I AM EXECUTING"<<endl
#define testcases int asdf; cin>>asdf; while(asdf--)
#define vi vector<int>
#define vll vector<long long int>
#define vppo(prs) for(auto x:prs){cout<<x.first<<" "<<x.second<<endl;}
#define For(__,hajmola,adfdf) for(int __ = hajmola; __<adfdf;__++)

string sconvert(ll int n)
{
	stringstream ss;
	ss<<n;
	string str = ss.str();
	return str;
}
bool sortbysec(const pair<int,int> &a, 
              const pair<int,int> &b) 
{ 
    return (a.second > b.second); 
}

template<typename T>
void imax(T &x,T y){
	x = max(x,y);
}
template<typename T>
void imin(T &x,T y){
	x = min(x,y);
}

void  single()
{
	
	ll int n;
	cin>>n;
    ll int m;
    cin>>m;
    vll a(n);
    ai(a,n);
    vll b(n);
    ai(b,n);
    vll ones;
    vll twos;
    for(int i=0;i<n;i++){
    	if(b[i]==2){
    		twos.pb(a[i]);
    	}
    	else{
    		ones.pb(a[i]);
    	}
    }
    sort(rall(ones));
    sort(rall(twos));
    // ao(ones);
    // cout<<endl;
    // ao(twos);
    // cout<<endl;
    vll sums;
    ll int sum=0;
    for(ll int i=0;i<ones.size();i++){
    	sum+=ones[i];
        sums.pb(sum);
    }
    sum=0;
    ll int answer=1e18;
    if(lower_bound(all(sums),m)!=sums.end()){
    ll int needed = lower_bound(all(sums),m)-sums.begin()+1;
    imin(answer,needed);
}
    for(ll int i=0;i<twos.size();i++){
         sum+=twos[i];
         ll int req = m-sum;
         if(req<=0){
         	imin(answer,(i+1)*2);
         }
         else{

         	auto it = lower_bound(all(sums),req);
         	if(it==sums.end()){
         		continue;
         	}
         	ll int needed = it-sums.begin()+1;
           // cout<<req<<" "<<needed<<endl;
         	imin(answer,(i+1)*2+needed);
         }
    }
    if(answer==1e18){
    	cout<<-1<<endl;
    	return;
    }
    cout<<answer<<endl;
 	return;
}
void multiple()
{
	testcases
    {
		single();
	}
}

int main()
{
  IOS;
 // freopen("input.txt","r",stdin);
  //freopen("output.txt","w",stdout);
  multiple();
}

















