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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int N = 5000, M = 100005;
vector<long long> g[N];
int a[N];
void solve() {
  long long i, j, n, m, k, a, b, sum = 0;
  cin >> n >> m;
  vector<long long> arr(n);
  for (i = 0; i < n; ++i) {
    cin >> arr[i];
    sum += arr[i];
  }
  vector<long long> v[3];
  for (i = 0; i < n; ++i) {
    cin >> k;
    v[k].push_back(arr[i]);
  }
  if (sum < m) {
    cout << -1 << "\n";
    return;
  }
  sort(v[1].begin(), v[1].end(), greater<int>());
  sort(v[2].begin(), v[2].end(), greater<int>());
  a = 0, b = 0;
  for (i = 1; i < v[1].size(); ++i) v[1][i] += v[1][i - 1];
  for (i = 1; i < v[2].size(); ++i) v[2][i] += v[2][i - 1];
  k = lower_bound(v[2].begin(), v[2].end(), m) - v[2].begin();
  long long ans = (k + 1) * 2;
  for (i = 0; i < v[1].size(); ++i) {
    k = m - v[1][i];
    if (k <= 0) {
      ans = min(ans, (i + 1));
      break;
    }
    j = lower_bound(v[2].begin(), v[2].end(), k) - v[2].begin();
    ans = min(ans, (i + 1) + j * 2 + 2);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int test;
  cin >> test;
  while (test--) {
    solve();
  }
}
int mpow(int base, int exp) {
  base %= ((int)1e9 + 7);
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % ((int)1e9 + 7);
    base = ((long long)base * base) % ((int)1e9 + 7);
    exp >>= 1;
  }
  return result;
}
void ipgraph(int n, int m) {
  int i, u, v;
  while (m--) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
void dfs(int u, int par) {
  for (int v : g[u]) {
    if (v == par) continue;
    dfs(v, u);
  }
}
