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;
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]" << endl; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
const int mod = 1e9 + 7;
const int N = 3e5 + 1;
void solve() {
  int n, m;
  cin >> n >> m;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> one, two;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x == 1)
      one.push_back(a[i]);
    else
      two.push_back(a[i]);
  }
  sort(one.begin(), one.end());
  sort(two.begin(), two.end());
  reverse(one.begin(), one.end());
  reverse(two.begin(), two.end());
  int j = 0, ans = 0;
  for (int i = 0; i < one.size(); i++) {
    if (one[i] >= m || j == two.size())
      m -= one[i], ans++;
    else if (two[j] >= m)
      m -= two[j++], ans += 2;
    else if (i + 1 < one.size() && one[i] + one[i + 1] >= m)
      m -= one[i] + one[i + 1], i++, ans += 2;
    else if (one[i] + two[j] >= m)
      m -= one[i] + two[j], ans += 3, j++;
    else if (i + 1 < one.size() && one[i] + one[i + 1] > two[j])
      m -= one[i] + one[i + 1], i++, ans += 2;
    else
      m -= two[j++], ans += 2, i--;
    if (m <= 0) break;
  }
  while (m > 0 && j < two.size()) m -= two[j++], ans += 2;
  if (m > 0) ans = -1;
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
