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 unsigned long long ull;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef double db;
#define all(x) (x).begin(), (x).end()

const ll MAXN = 1e9, MINN = -1e9;
const ld pi = 3.14159265358979323846264338327950288;

void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
}

ll gcd(ll a, ll b) {
    while (b) {
        a %= b;
        swap(a, b);
    }
    return a;
}

void solve() {
    ll n, m;
    cin >> n >> m;
    vector<ll> a(n);
    vector<ll> b(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) cin >> b[i];
    vector<ll> one, two;
    for (int i = 0; i < n; i++) {
        if (b[i] == 1) {
            one.push_back(a[i]);
        } else {
            two.push_back(a[i]);
        }
    }
    sort(all(one));
    sort(all(two));
    one.push_back(0);
    two.push_back(0);
    reverse(all(one));
    reverse(all(two));

    vector<ll> pref1(one.size()), pref2(two.size());

    for (int i = 1; i < one.size(); i++) {
        pref1[i] = pref1[i - 1] + one[i];
    }
    for (int i = 1; i < two.size(); i++) {
        pref2[i] = pref2[i - 1] + two[i];
    }
    ll ans = 1e18;
//    for (auto i : one) cout << i << " ";
//    cout << '\n';
//    for (auto i : two) cout << i << " ";
//    cout << "\n";
    for (int i = 0; i < one.size(); i++) {
        int now = i;
        ll find = m - pref1[i];
        auto it = lower_bound(all(pref2), find);
        if (it == pref2.end()) continue;
        ll cnt = it - pref2.begin();
        ll cur = now + cnt * 2;
        ans = min(ans, cur);
    }
    for (int i = 0; i < two.size(); i++) {
        int now = i;
        ll find = m - pref2[i];
        auto it = lower_bound(all(pref1), find);
        if (it == pref1.end()) continue;
        ll cnt = it - pref1.begin();
        ll cur = cnt + now * 2;
        ans = min(ans, cur);
    }
    if (ans == 1e18) {
        cout << -1 << "\n";
    } else {
        cout << ans << "\n";
    }
}

int main() {
    ll t = 1;
    fastio();
    cin >> t;
    while (t--) { solve(); }
    return 0;
}