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 ll long long
#define PI 3.14159265
ll mod = 1000000007;
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
int main()  
{
    #ifndef ONLINE_JUDGE
        freopen("input1.txt","r",stdin);
        freopen("output1.txt","w",stdout);
    #endif
    ios_base::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    ll t;
    cin >> t;
    while(t--)  {
        ll n, m;
        cin >> n >> m;
        vector<ll> one, two;
        vector<ll> v1(n), v2(n);
        ll sum = 0;
        for(int i = 0; i < n; i++)  {
            cin >> v1[i];
            sum += v1[i];
        }
        for(int i = 0; i < n; i++)  {
            cin >> v2[i];
            if(v2[i] == 1)  {
                one.push_back(v1[i]);
            }
            else    {
                two.push_back(v1[i]);
            }
        }
        sort(one.rbegin(), one.rend());
        sort(two.rbegin(), two.rend());
        if(sum < m) {
            cout << "-1" << "\n";
            continue;
        }
        ll idx = 0;
        ll sz1 = one.size(), sz2 = two.size();
        sum = 0;
        while(idx < sz1 && sum < m) {
            sum += one[idx];
            idx++;
        }
        ll ans = 3*n;
        if(sum >= m)    {
            ans = idx;
        }
        ll idx1 = idx, idx2 = 0;
        //reverse(one.begin(), one.end());
        while(idx2 < sz2)   {
            sum += two[idx2];
            idx2++;
            while(idx1 > 0 && sum - one[idx1 - 1] >= m)    {
                sum -= one[idx1 - 1];
                idx1--;
            }
            if(sum >= m)    {
                ans = min(ans, idx2*2 + idx1);
            }
        }
        if(ans > 2*n)   {
            cout << "-1" << "\n";
        }
        else    {
            cout << ans << "\n";
        }
    }
}
    
    

