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 itval(istream_iterator<string> it) {}
template <typename T, typename... Args>
void itval(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  itval(++it, args...);
}
const long long int MOD = 1e9 + 7;
template <typename T>
inline void print(T x) {
  cout << x << "\n";
}
template <typename T>
inline void printvec(T x) {
  for (auto a : x) cout << a << ' ';
  cout << '\n';
}
struct custom {
  bool operator()(
      const pair<long long int, pair<long long int, long long int> > &p1,
      const pair<long long int, pair<long long int, long long int> > &p2)
      const {
    return p1.first < p2.first;
  }
};
long long int get_pow(long long int a, long long int b, long long int M = MOD) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % M;
    a = (a * a) % M;
    b >>= 1;
  }
  return res;
}
const long long int N = 1e5 + 7, inf = 2e18;
void solve() {
  long long int n, m, sum = 0, x;
  cin >> n >> m;
  vector<long long> a(n), one, two;
  for (long long int i = (long long int)0; i < (long long int)(n); i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (long long int i = (long long int)0; i < (long long int)(n); i++) {
    cin >> x;
    if (x == 1)
      one.push_back(a[i]);
    else
      two.push_back(a[i]);
  }
  if (sum < m) {
    cout << "-1\n";
    return;
  }
  sort(one.begin(), one.end(), greater<long long int>());
  sort(two.begin(), two.end(), greater<long long int>());
  long long int ans = 0, oa = one.size(), ta = two.size(), u1 = 0, u2 = 0,
                tot = m;
  while (u1 < oa && u2 < ta && m > 0) {
    if (one[u1] > two[u2] || one[u1] >= m) {
      m -= one[u1];
      u1++;
      ans++;
    } else if (u1 + 1 < oa && one[u1] + one[u1 + 1] >= m) {
      m -= one[u1];
      m -= one[u1 + 1];
      u1 += 2;
      ans += 2;
    } else if (two[u2] >= m) {
      m -= two[u2];
      u2++;
      ans += 2;
    } else if (one[u1] + two[u2] >= m) {
      m -= one[u1];
      m -= two[u2];
      u1++;
      u2++;
      ans += 3;
    } else if (u1 + 1 < oa && one[u1] + one[u1 + 1] >= two[u2]) {
      m -= one[u1];
      m -= one[u1 + 1];
      u1 += 2;
      ans += 2;
    } else {
      m -= two[u2];
      u2++;
      ans += 2;
    }
  }
  if (m <= 0) {
    cout << ans << '\n';
    return;
  }
  while (u1 < oa && m > 0) {
    ans++;
    m -= one[u1];
    u1++;
  }
  while (u2 < ta && m > 0) {
    ans += 2;
    m -= two[u2];
    u2++;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  cin >> test;
  clock_t z = clock();
  for (long long int tes = (long long int)0; tes < (long long int)(test);
       tes++) {
    solve();
  }
  fprintf(stderr, "Total Time:%.4f\n", (double)(clock() - z) / CLOCKS_PER_SEC),
      fflush(stderr);
  return 0;
}
