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;
long long int exp(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long int modinv(long long int x) { return exp(x, 1000000007 - 2); }
long long int add(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  a = ((a + b) % 1000000007 + 1000000007) % 1000000007;
  return a;
}
long long int sub(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  a = ((a - b) % 1000000007 + 1000000007) % 1000000007;
  return a;
}
long long int mul(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  a = ((a * b) % 1000000007 + 1000000007) % 1000000007;
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int Log2n(long long int n) { return (n > 1) ? 1 + Log2n(n / 2) : 0; }
long long int fac[100009];
long long int ncr_mod(long long int n, long long int k) {
  long long int ans = fac[n];
  ans *= modinv(fac[k]);
  ans %= 1000000007;
  ans *= modinv(fac[n - k]);
  ans %= 1000000007;
  return ans;
}
vector<long long int> v_prime;
void Sieve(long long int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (prime[p]) v_prime.push_back(p);
}
vector<long long int> v_factor;
void factors(long long int n) {
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        v_factor.push_back(i);
      else {
        v_factor.push_back(i), v_factor.push_back(n / i);
      };
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  ;
  long long int t = 1;
  cin >> t;
  while (t--) {
    long long int n, i, x, y, ok = 0, sum = 0, ans = 0, j, k, cnt = 0, m, c = 0,
                              sum1 = 0, res = 0;
    long long int h[100009] = {0};
    cin >> n >> m;
    string s;
    vector<long long int> a(n, 0), v, b(n, 0), v1, v2;
    multiset<long long int> s1, s2;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    for (i = 0; i < n; i++) {
      cin >> b[i];
      sum1 += b[i];
      if (b[i] == 1)
        s1.insert(a[i]);
      else
        s2.insert(a[i]);
    }
    if (m > sum) {
      cout << -1 << "\n";
      continue;
    }
    if (m == sum) {
      cout << sum1 << "\n";
      continue;
    }
    k = m;
    sum = 0, sum1 = 0;
    long long int ext = 0, sum2 = 0, sum12 = 0, sum3 = 0, sum4 = 0;
    while (1) {
      if (s1.size() > 0) {
        auto it = s1.end();
        it--;
        x = *it;
        if (x >= k) {
          k = 0;
          sum += 1;
          break;
        }
      }
      if (s1.size() > 0 && s2.size() > 0) {
        auto it1 = s1.end();
        it1--;
        sum3 = *it1;
        if (sum3 >= k) {
          sum += 1;
          k = 0;
          break;
        }
        if (s1.size() >= 2) {
          auto it19 = s1.end();
          it19--;
          it19--;
          sum3 += *it19;
          if (sum3 >= k) {
            sum += 2;
            break;
          } else {
            sum3 = sum3 - *it19;
          }
        }
        auto it2 = s2.end();
        it2--;
        sum3 += *it2;
        sum4 = *it2;
        if (k <= sum4) {
          k = 0;
          sum += 2;
          break;
        } else if (k <= sum3) {
          k = 0;
          sum += 3;
          break;
        }
      }
      if (s1.size() >= 2) {
        auto it1 = s1.end();
        it1--;
        sum1 = *it1;
        sum12 = sum1;
        it1--;
        sum1 += *it1;
        if (s2.size() > 0) {
          auto it2 = s2.end();
          it2--;
          sum2 = *it2;
          if (max(sum1, sum2) == sum1) {
            auto it1 = s1.end();
            it1--;
            long long int x1 = *it1;
            s1.erase(it1);
            it1 = s1.end();
            it1--;
            k = k - sum1;
            sum = sum + 2;
            s1.erase(it1);
          } else {
            k = k - sum2;
            sum = sum + 2;
            auto it1 = s2.end();
            it1--;
            s2.erase(it1);
          }
        } else {
          auto it1 = s1.end();
          it1--;
          s1.erase(it1);
          it1 = s1.end();
          it1--;
          s1.erase(it1);
          k = k - sum1;
          sum = sum + 2;
        }
      } else {
        if (s1.size() == 0) {
          auto it2 = s2.end();
          it2--;
          sum2 = *it2;
          k = k - sum2;
          if (ok == 1 && k < 0) {
            if (k + ext < 0) sum--;
          }
          sum += 2;
        } else {
          sum1 = *s1.begin();
          if (k - sum1 <= 0) {
            k = 0;
            sum += 1;
          } else {
            ok = 1;
            ext = *s1.begin();
            k = k - ext;
            s1.erase(s1.begin());
          }
        }
      }
      if (k <= 0) break;
    }
    cout << sum << "\n";
  }
}
