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 int N = 2e5 + 10;
int n, m, cas, cnt1, cnt2;
int tmp[N], opt[N], a[N], b[N];
bool cmp(int a, int b) { return a > b; }
int main() {
  scanf("%d", &cas);
  while (cas--) {
    scanf("%d%d", &n, &m);
    long long ss = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &tmp[i]);
    for (int i = 1; i <= n; i++) scanf("%d", &opt[i]);
    cnt1 = cnt2 = 0;
    for (int i = 1; i <= n; i++) {
      if (opt[i] == 1)
        a[++cnt1] = tmp[i];
      else
        b[++cnt2] = tmp[i];
      ss += tmp[i];
    }
    if (ss < m) {
      printf("-1\n");
      continue;
    }
    sort(a + 1, a + cnt1 + 1, cmp);
    sort(b + 1, b + cnt2 + 1, cmp);
    long long sum1 = 0;
    int r = 1;
    for (r = 1; r <= cnt1; r++) {
      sum1 += a[r];
      if (sum1 >= m) {
        break;
      }
    }
    long long sum2 = 0;
    long long ans = 1e9;
    b[0] = 0;
    for (int y = 0; y <= cnt2; y++) {
      sum2 += b[y];
      while (r >= 1 && sum2 + sum1 - a[r] >= m) {
        sum1 -= a[r];
        r--;
      }
      if (sum1 + sum2 >= m) ans = min(ans, 1ll * r + 2 * y);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
