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 MOD = 1e9 + 7;
const long long INF = 1e9 + 7;
long long int nCrModpDP(long long int n, long long int r, int p) {
  long long int C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
long long int nCrModpLucas(long long int n, long long int r, long long int p) {
  if (r == 0) return 1;
  long long int ni = n % p, ri = r % p;
  return (nCrModpLucas(n / p, r / p, p) * nCrModpDP(ni, ri, p)) % p;
}
int main() {
  long long int t, n, m;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    vector<long long int> v1, v2;
    long long int arr[n];
    for (int i = 0; i < n; i++) cin >> arr[i];
    long long int num, sumB = 0;
    for (int i = 0; i < n; i++) {
      cin >> num;
      if (num == 1)
        v1.push_back(arr[i]);
      else {
        v2.push_back(arr[i]);
        sumB += arr[i];
      }
    }
    sort(v1.begin(), v1.end(), greater<long long int>());
    sort(v2.begin(), v2.end(), greater<long long int>());
    long long int sum = 0, ans = INT_MAX;
    int second = v2.size();
    bool f = false, f1, f11;
    int j = second - 1;
    f11 = false;
    while (j >= 0 && sum + sumB > m) {
      sumB -= v2[j];
      j--;
      f11 = true;
    }
    if (sum + sumB == m) {
      ans = 2 * (j + 1);
      f = true;
    }
    if (sum + sumB < m && f11) {
      ans = 2 * (j + 2);
      j++;
      sumB += v2[j];
      f = true;
    }
    long long int cnt = 0;
    for (int i = 0; i < v1.size(); i++) {
      cnt = i + 1;
      sum += v1[i];
      f1 = false;
      while (j >= 0 && sum + sumB > m) {
        sumB -= v2[j];
        j--;
        f1 = true;
      }
      if (sum + sumB > m && f1) {
        ans = min(ans, cnt);
        f = true;
      }
      if (sum + sumB == m && !f1) {
        cnt += 2 * (j + 1);
        ans = min(ans, cnt);
        f = true;
      }
      if (sum + sumB < m && f1) {
        cnt += 2 * (j + 2);
        j++;
        sumB += v2[j];
        ans = min(ans, cnt);
        f = true;
      } else if (sum + sumB == m && f1) {
        cnt += 2 * (j + 1);
        ans = min(ans, cnt);
        f = true;
      } else if (sum >= m && !f1) {
        ans = min(ans, cnt);
        f = true;
      }
      if (j < 0) break;
    }
    if (f)
      cout << ans << endl;
    else
      cout << "-1" << endl;
  }
  return 0;
}
