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;
vector<long long> b1, b2;
long long func(long long n) {
  long long ans = 0;
  queue<long long> q1, q2;
  for (long long x = 0; x <= (signed)b1.size() - 1; x++) q1.push(b1[x]);
  for (long long x = 0; x <= (signed)b2.size() - 1; x++) q2.push(b2[x]);
  long long min1 = LONG_LONG_MAX, min2 = LONG_LONG_MAX;
  while (n > 0) {
    if (q1.empty() && q2.empty()) break;
    if (q1.empty() && !q2.empty()) {
      if (n >= 2) {
        n = n - 2;
        ans = ans + q2.front();
        min2 = min(min2, q2.front());
        q2.pop();
      } else {
        break;
      }
      continue;
    }
    if (q2.empty() && !q1.empty()) {
      if (n >= 1) {
        n = n - 1;
        ans = ans + q1.front();
        min1 = min(min1, q1.front());
        q1.pop();
      } else {
        break;
      }
      continue;
    }
    if (n == 1) {
      if (!q1.empty()) {
        ans = ans + q1.front();
        min1 = min(min1, q1.front());
        q1.pop();
      }
      break;
    }
    if (q2.front() >= 2 * q1.front()) {
      n = n - 2;
      ans = ans + q2.front();
      min2 = min(min2, q2.front());
      q2.pop();
    }
    if (q2.front() < 2 * q1.front()) {
      n = n - 1;
      ans = ans + q1.front();
      min1 = min(min1, q1.front());
      q1.pop();
    }
  }
  long long t = ans;
  if (n == 1) {
    if (!q2.empty()) {
      ans = ans - min1;
      ans = ans + q2.front();
    }
  }
  return max(t, ans);
}
signed main() {
  long long T;
  cin >> T;
  while (T--) {
    b1.clear();
    b2.clear();
    long long n, m;
    cin >> n >> m;
    long long a[n + 5], b[n + 5];
    long long cost = 0, sum = 0;
    for (long long x = 1; x <= n; x++) cin >> a[x];
    for (long long x = 1; x <= n; x++) {
      long long t;
      cin >> t;
      if (t == 1)
        b1.push_back(a[x]);
      else
        b2.push_back(a[x]);
      cost += t;
      sum += a[x];
    }
    sort(b1.begin(), b1.end(), greater<long long>());
    sort(b2.begin(), b2.end(), greater<long long>());
    if (sum < m) {
      cout << "-1\n";
      continue;
    }
    long long low = 1, high = cost;
    long long ans = INT_MAX;
    while (low <= high) {
      long long mid = (low + high) / 2;
      long long a = func(mid);
      if (a >= m) {
        ans = min(ans, mid);
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    cout << ans << "\n";
  }
}
