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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long Mod = 1e9 + 7;
const long long inf = 1e18;
const long long N = 2e5;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    vector<pair<long long, long long>> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i].first;
    }
    for (long long i = 0; i < n; i++) {
      cin >> a[i].second;
    }
    vector<long long> one, two;
    for (auto [i, j] : a) {
      if (j == 1) {
        one.push_back(i);
      } else {
        two.push_back(i);
      }
    }
    sort(one.rbegin(), one.rend());
    sort(two.rbegin(), two.rend());
    debug() << "\033[0;33mLine(" << 112 << ")\033[0m"
            << "\033[1;36m->"
            << "\033[0m"
            << "\033[94m ["
            << "one"
               ": "
            << (one) << "]\033[0m";
    debug() << "\033[0;33mLine(" << 113 << ")\033[0m"
            << "\033[1;36m->"
            << "\033[0m"
            << "\033[94m ["
            << "two"
               ": "
            << (two) << "]\033[0m";
    long long sum = 0;
    for (long long i = 0; i < (long long)one.size(); i++) {
      if (i > 0) {
        one[i] += one[i - 1];
      }
    }
    sum = 0;
    for (long long i = 0; i < (long long)two.size(); i++) {
      if (i > 0) {
        two[i] += two[i - 1];
      }
    }
    long long ans = inf;
    for (long long i = 0; i < (long long)one.size(); i++) {
      long long taken = one[i];
      long long left = max(0ll, (m - taken));
      long long l = lower_bound(two.begin(), two.end(), left) - two.begin();
      if (l != (long long)two.size()) {
        long long cost = i + 1 + 2 * (l + 1);
        ans = min(ans, cost);
      }
    }
    for (long long i = 0; i < (long long)two.size(); i++) {
      long long taken = two[i];
      long long left = max(0ll, m - taken);
      long long l = lower_bound(one.begin(), one.end(), left) - one.begin();
      if (l != (long long)one.size()) {
        long long cost = 2 * (i + 1) + (l + 1);
        ans = min(ans, cost);
      }
    }
    if (ans == inf) ans = -1;
    cout << ans << '\n';
  }
}
