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;
const double pi = acos(-1.0);
const int mod = (1 ? 1000000007 : 998244353);
const int inf = (1 ? 0x3f3f3f3f : 0x7fffffff);
const int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
const long long INF = (1 ? 0x3f3f3f3f3f3f3f3f : 0x7fffffffffffffff);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long poww(long long a, long long b) {
  long long s = 1;
  while (b) {
    if (b & 1) s = (s * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return s % mod;
}
const int N = 2e5 + 11;
long long n, m;
int z[N];
void solve() {
  vector<int> ve1, ve2;
  cin >> n >> m;
  int now;
  long long sum = 0;
  for (int i = 0; i < n; i++) cin >> z[i];
  for (int i = 0; i < n; i++) {
    cin >> now;
    if (now == 1)
      ve1.push_back(z[i]);
    else
      ve2.push_back(z[i]);
  }
  sort(ve1.begin(), ve1.end());
  sort(ve2.begin(), ve2.end());
  int size1 = ve1.size();
  int size2 = ve2.size();
  int times = 0;
  while (sum < m) {
    if (size1 == 0 && size2 == 0) {
      cout << -1 << "\n";
      return;
    }
    if (size2 == 0) {
      size1--;
      sum += ve1[size1];
      times++;
    } else if (size1 == 0) {
      size2--;
      sum += ve2[size2];
      times += 2;
    } else if (sum + ve1[size1 - 1] >= m) {
      cout << times + 1 << "\n";
      return;
    } else if (size1 == 1) {
      if (ve1[size1 - 1] >= ve2[size2 - 1]) {
        size1--;
        sum += ve1[size1];
        times++;
      } else {
        size2--;
        sum += ve2[size2];
        times += 2;
      }
    } else if (sum + ve1[size1 - 1] + ve1[size1 - 2] >= m) {
      cout << times + 2 << "\n";
      return;
    } else if (sum + ve2[size2 - 1] >= m) {
      cout << times + 2 << "\n";
      return;
    } else if (sum + ve1[size1 - 1] + ve2[size2 - 1] >= m) {
      cout << times + 3 << "\n";
      return;
    } else {
      if (ve1[size1 - 1] + ve1[size1 - 2] >= ve2[size2 - 1]) {
        size1--;
        sum += ve1[size1];
        size1--;
        sum += ve1[size1];
        times += 2;
      } else {
        size2--;
        sum += ve2[size2];
        times += 2;
      }
    }
  }
  cout << times << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return 0;
}
