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 long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const long double EPS = 1e-9;
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline double my_sqrt(T n) {
  double high = n, low = 0, mid, ans;
  int cnt = 100;
  while (cnt--) {
    mid = low + (high - low) / 2;
    if (mid * mid <= n)
      ans = mid, low = mid;
    else
      high = mid;
  }
  return ans;
}
template <class T>
inline T bigmod(T b, T p) {
  if (p <= 0 || b == 0) return 1;
  if (p & 1) return (b * bigmod(b, p - 1)) % MOD;
  long long x = bigmod(b, p >> 1);
  return (x * x) % MOD;
}
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  std::vector<int> a, b, v(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    int move;
    scanf("%d", &move);
    if (move == 1)
      a.push_back(v[i]), sum += v[i];
    else
      b.push_back(v[i]);
  }
  sort((a).begin(), (a).end());
  sort((b).begin(), (b).end());
  int taken = a.size();
  int indx = 0;
  while (indx < a.size() && sum - a[indx] >= m) {
    sum -= a[indx];
    taken--;
    indx++;
  }
  int indx2 = b.size() - 1;
  while (indx2 >= 0 && sum < m) {
    sum += b[indx2--];
    taken += 2;
  }
  while (indx < a.size() && sum - a[indx] >= m) {
    sum -= a[indx];
    taken--;
    indx++;
  }
  int f = 1;
  long long ex;
  while (a.size() - indx >= 2 && indx2 >= 0 && sum >= m) {
    if (!f)
      ex = sum - m + b[indx2];
    else
      f = 0, ex = b[indx2];
    if (ex >= a[indx] + a[indx + 1]) {
      sum += b[indx2];
      taken += 2;
      while (indx < a.size() && sum - a[indx] >= m) {
        sum -= a[indx];
        taken--;
        indx++;
      }
      indx2--;
      indx++;
    } else
      break;
  }
  if (sum >= m) {
    printf("%d\n", taken);
  } else
    puts("-1");
  return;
}
int main() {
  int test = 1;
  scanf("%d", &test);
  while (test--) {
    solve();
  }
  return 0;
}
