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;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vii;
typedef vector<vi> vvi;
typedef vector<pll> vll;
typedef vector<vl> vvl;

#define fori(i, n) for (int i = 0; i < n; i++)
#define ford(i, n) for (int i = n - 1; i >= 0; i--)
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define repd(i, a, b) for (int i = a; i >= b; i--)
#define trav(x, a) for (auto &x : a)
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define eb emplace_back
#define endl '\n'
#define sz(a) (int)(a).size()
#define fi first
#define se second

clock_t time_p = clock();
void time_taken()
{
    time_p = clock() - time_p;
    cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}

const ll mod = 1e9 + 7;
const ll INF = 1e18;

void solve()
{
    int n, m;
    cin >> n >> m;
    vi a(n), b(n);
    fori(i, n) cin >> a[i];
    fori(i, n) cin >> b[i];
    vl adj[2];
    fori(i, n)
    {
        --b[i];
        adj[b[i]].pb(a[i]);
    }
    sort(all(adj[0]));
    sort(all(adj[1]));
    reverse(all(adj[0]));
    reverse(all(adj[1]));

    fori(i, 2)
    {
        auto &v = adj[i];
        fori(i, sz(v))
        {
            if (i)
                v[i] += v[i - 1];
        }
    }

    int lo = 0, hi = 2 * n;
    int ans = -1;
    while (lo <= hi)
    {
        int mid = (lo + hi) / 2;
        bool ok = 0;
        for (int i = 0; i <= min(mid, sz(adj[0])); i++)
        {
            int rem = mid - i;
            int can = rem / 2;
            can = min(can, sz(adj[1]));
            if ((i ? adj[0][i - 1] : 0) + (can ? adj[1][can - 1] : 0) >= m)
            {
                ok = 1;
                break;
            }
        }
        if (ok)
        {
            hi = mid - 1;
            ans = mid;
        }
        else
        {
            lo = mid + 1;
        }
    }
    cout << ans << endl;
}

int main()
{
    ios_base::sync_with_stdio(false), cin.tie(nullptr);

    int t;
    cin >> t;
    while (t--)
        solve();

    time_taken();
    return 0;
}