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;
vector<int> final;
void seive(int n) {
  vector<bool> prime(n + 1, true);
  prime[0] = false;
  prime[1] = false;
  for (long long int i = 2; i <= (long long int)sqrt(n); ++i) {
    if (prime[i]) {
      for (int j = i * i; j <= n; j += i) prime[j] = false;
    }
  }
  for (long long int i = 2; i <= n; ++i) {
    if (prime[i]) final.push_back(i);
  }
}
const int mod = 998244353;
const int mx = 2e5 + 1;
vector<int> fact(mx);
vector<int> inv_fact(mx);
long long int expo_pow(long long int a, long long int b) {
  a %= mod;
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
void precompute() {
  fact[0] = 1;
  for (long long int i = 1; i <= mx; ++i) fact[i] = (fact[i - 1] * i) % mod;
  for (long long int i = 0; i <= mx; ++i)
    inv_fact[i] = expo_pow(fact[i], mod - 2) % mod;
}
long long int compute(long long int n, long long int r) {
  return fact[n] * 1ll * inv_fact[r] % mod * inv_fact[n - r] % mod;
}
long long int single(vector<long long int>& a, long long int m) {
  long long int i = 0, sum = 0;
  while (i < a.size()) {
    sum += a[i];
    if (sum >= m) return i + 1;
    i++;
  }
  return -1;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n, m, x;
    cin >> n >> m;
    vector<long long int> a(n), ones, twos;
    for (long long int i = 0; i < n; ++i) cin >> a[i];
    for (long long int i = 0; i < n; ++i) {
      cin >> x;
      if (x == 1)
        ones.push_back(a[i]);
      else
        twos.push_back(a[i]);
    }
    long long int i = 0, sum = 0, val = 0, j = 0;
    if (ones.size()) {
      sort(ones.begin(), ones.end(), greater<long long int>());
      if (twos.size()) {
        long long int currval = (long long int)1e18;
        sort(twos.begin(), twos.end(), greater<long long int>());
        vector<long long int> m1, m2;
        for (long long int i = 0; i < ones.size(); ++i) {
          sum += ones[i];
          if (sum >= m) currval = min(currval, i + 1);
          m1.push_back(sum);
        }
        sum = 0;
        for (long long int i = 0; i < twos.size(); ++i) {
          sum += twos[i];
          if (sum >= m) currval = min(currval, 2 * (i + 1));
          m2.push_back(sum);
        }
        if (m1[ones.size() - 1] + m2[twos.size() - 1] >= m) {
          for (long long int i = 0; i < ones.size(); ++i) {
            if (m - m1[i] > 0) {
              auto pos =
                  lower_bound(m2.begin(), m2.end(), m - m1[i]) - m2.begin();
              if (pos == m2.size())
                continue;
              else
                currval = min(currval, (i + 1) + 2 * (pos + 1));
            }
          }
          for (long long int i = 0; i < twos.size(); ++i) {
            if (m - m2[i] > 0) {
              auto pos =
                  lower_bound(m1.begin(), m1.end(), m - m2[i]) - m1.begin();
              if (pos == m1.size())
                continue;
              else
                currval = min(currval, 2 * (i + 1) + (pos + 1));
            }
          }
          val = currval;
        } else
          val = -1;
      } else
        val = single(ones, m);
    } else {
      sort(twos.begin(), twos.end(), greater<long long int>());
      val = 2 * single(twos, m);
    }
    cout << val << "\n";
  }
}
