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>
using namespace std;
#define endl "\n"
#define ll long long
#define ld long double
const ll MOD=998244353;
const ll LMAX=1e18;
const ll MAX=2e5+5;

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

int main() {
#ifndef ONLINE_JUDGE
  	//for getting input from input.txt
   	freopen("input.txt","r",stdin);
   	//for writing output  to output.txt
   	freopen("output.txt","w",stdout);
#endif

    ios_base::sync_with_stdio();
    cin.tie();
    cout.tie();
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,m,sum_1=0,sum_2=0;
        cin>>n>>m;
        vector<ll> a(n);
        vector<ll> one;
        vector<ll> two;

        for(ll i=0;i<n;i++)
        {
            cin>>a[i];
        }
        for(ll i=0;i<n;i++)
        {
            ll b;
            cin>>b;
            if(b==1)
            {
                one.push_back(a[i]);
            }
            else
            {
                two.push_back(a[i]);
                sum_2+=a[i];
            }
        }
        sort(one.rbegin(),one.rend());
        sort(two.rbegin(),two.rend());
        ll r=(ll)two.size()-1;
        ll ans=INT_MAX;

        for(ll i=0;i<=one.size();i++)
        {
            while(r>=0 && (sum_1+sum_2-two[r])>=m)
            {
                sum_2=sum_2-two[r];
                r--;
            }
            if((sum_1+sum_2)>=m)
            {
                ans=min(ans,2*(r+1)+i);
                //cout<<i<<": "<<ans<<" ";
            }
            if(i==one.size())
            {
                break;
            }
            sum_1=sum_1+one[i];
        }

        cout<<(ans==INT_MAX ? -1:ans)<<endl;
    }
#ifdef _DEBUG
    cerr << "TIME = " << clock() — tt << endl;
    tt = clock();
#endif
    return 0;
}
    /*taskkill -im .exe -f*/

