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 MOD = 1e9;
const int maxn = 1e6 + 5;
int n, m;
pair<int, int> a[200005];
int f[200005];
bool Check(int mid) {
  int cur = 0;
  long long sum = 0;
  vector<int> v;
  for (int i = n; i >= 1; --i) {
    if (cur + a[i].second <= mid) {
      if (a[i].second == 2) v.push_back(a[i].first);
      sum += a[i].first;
      cur += a[i].second;
    }
  }
  int cnt = 0;
  int s = 0;
  int x = 0;
  for (int i = n; i >= 1; --i) {
    if (cur + a[i].second <= mid) {
      cur += a[i].second;
    } else {
      if (a[i].second == 1) {
        s += a[i].first;
        ++cnt;
        if (cnt == 2) {
          while (x < (int)v.size() && v[x] >= s) {
            ++x;
          }
          if (x < (int)v.size()) {
            sum -= v[x];
            sum += s;
          }
          cnt = 0;
        }
      }
    }
  }
  if (sum >= m) return 1;
  return 0;
}
void Solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].second;
  }
  int l = 0, r = 2 * n + 1;
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= n; ++i) {
    if (a[i].second == 1)
      f[i] = a[i].first;
    else
      f[i] = f[i - 1];
  }
  while (r - l > 1) {
    int mid = (l + r) >> 1;
    if (Check(mid))
      r = mid;
    else
      l = mid;
  }
  if (r == 2 * n + 1) r = -1;
  cout << r << "\n";
}
int res = 1e9 + 7;
void Cal(int mask) {
  long long s = 0;
  int cur = 0;
  for (int i = 1; i <= n; ++i)
    if (((mask >> (i - 1)) & 1)) {
      s += a[i].first;
      cur += a[i].second;
    }
  if (s >= m) res = min(res, cur);
}
void BF() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].first;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> a[i].second;
  }
  res = 1e9 + 7;
  for (int mask = 0; mask < (1 << n); ++mask) Cal(mask);
  if (res == 1e9 + 7) res = -1;
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  if (fopen(""
            ".inp",
            "r")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  int test_case = 1;
  cin >> test_case;
  while (test_case--) {
    BF();
  }
  return 0;
}
