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;
int T, n, m;
int a[200005], b[200005];
int arr1[200005], arr2[200005];
long long sum1[200005], sum2[200005];
int cnt1, cnt2;
int read() {
  int sum = 0;
  char ch;
  do ch = getchar();
  while (ch < '0' || ch > '9');
  while (ch >= '0' && ch <= '9')
    sum = (sum << 1) + (sum << 3) + (ch ^ 48), ch = getchar();
  return sum;
}
bool cmp(int x, int y) { return x > y; }
int main() {
  cin >> T;
  while (T--) {
    scanf("%d%d", &n, &m);
    int i, j;
    long long sum = 0;
    for (i = 1; i <= n; i++) {
      a[i] = read();
      sum += a[i];
    }
    for (i = 1; i <= n; i++) {
      b[i] = read();
      if (b[i] == 1)
        arr1[++cnt1] = a[i];
      else
        arr2[++cnt2] = a[i];
    }
    if (sum < m) {
      cout << -1 << endl;
      continue;
    }
    sort(arr1 + 1, arr1 + cnt1 + 1, cmp);
    sort(arr2 + 1, arr2 + cnt2 + 1, cmp);
    for (i = 1; i <= cnt1; i++) sum1[i] = sum1[i - 1] + arr1[i];
    for (i = 1; i <= cnt2; i++) sum2[i] = sum2[i - 1] + arr2[i];
    int ans = 0x3f3f3f3f;
    int id = lower_bound(sum1 + 1, sum1 + cnt1 + 1, m) - sum1;
    if (id != cnt1 + 1) ans = min(ans, id);
    for (i = 0; i <= min(cnt1, id); i++) {
      int pos = lower_bound(sum2 + 1, sum2 + cnt2 + 1, m - sum1[i]) - sum2;
      if (pos != cnt2 + 1) ans = min(ans, i + 2 * pos);
    }
    cout << ans << endl;
    for (i = 1; i <= cnt1; i++) arr1[i] = 0, sum1[i] = 0;
    for (i = 1; i <= cnt2; i++) arr2[i] = 0, sum2[i] = 0;
    cnt1 = cnt2 = 0;
  }
  return 0;
}
