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 ull unsigned long long
#define pb push_back
using namespace std;
const int sz = 1e6 + 9;
const ll mod =  1e9 + 7;
const ll inf = 1e18;
const double eps = 1e-9;
void test()
{
    int n,m;cin>>n>>m;
    ll x[n],y[n];
    ll sum = 0;
    vector<ll> twoes,ones;
    for(int i=0;i<n;i++)
    {
        cin>>x[i];
        sum+=x[i];
    }
    for(int i=0;i<n;i++)
    {
        cin>>y[i];
        if(y[i]==1)
            ones.pb(x[i]);
        else
            twoes.pb(x[i]);
    }
    if(sum < m) {
        cout << "-1\n";
        return;
    }
    ll ans = 0;
    sum = m;
    sort(ones.rbegin(),ones.rend());
    sort(twoes.rbegin(),twoes.rend());
    int i = 0,j = 0;
    if(ones.empty())
    {
        while(sum > 0)
            sum-=twoes[i++],ans+=2;
        cout<<ans<<"\n";return;
    }
    else if(twoes.empty())
    {
        while(sum > 0)
            sum-=ones[i++],ans++;
        cout<<ans<<"\n";return;
        return;
    }
    while(sum > 0)
    {
        if(sum <= ones[j] && j < ones.size())
        {
            ans++;break;
        }
        if(twoes[i] >= ones[j] + ones[j + 1] && i < twoes.size() && j < ones.size() - 1)
        {
            sum-=twoes[i++],ans+=2;
        }
        else if(i < twoes.size() && j < ones.size() - 1)
            sum -= ones[j++],ans++;
        else if(i < twoes.size() && j==ones.size() - 1)
        {
            if(ones[j] > twoes[i])
                sum-=ones[j++],ans++;
            else
                sum-=twoes[i++],ans+=2;
        }
        else if(j < ones.size() && i==twoes.size())
        {
            while(sum > 0 && j < ones.size()) { sum -= ones[j++], ans++; }
        }
        else if(i < twoes.size() && j==ones.size())
        {
            while(sum > 0 && i < twoes.size()) { sum -= twoes[i++], ans += 2; }
        }
    }
    cout<<ans<<"\n";
}

int main(){
    int cases = 1;
    cin>>cases;
    while(cases--)
        test();
}