Consider some positive integer x. Its prime factorization will be of form x = 2^{k_1} ⋅ 3^{k_2} ⋅ 5^{k_3} ⋅ ...

Let's call x elegant if the greatest common divisor of the sequence k_1, k_2, ... is equal to 1. For example, numbers 5 = 5^1, 12 = 2^2 ⋅ 3, 72 = 2^3 ⋅ 3^2 are elegant and numbers 8 = 2^3 (GCD = 3), 2500 = 2^2 ⋅ 5^4 (GCD = 2) are not.

Count the number of elegant integers from 2 to n.

Each testcase contains several values of n, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^5) — the number of values of n in the testcase.

Each of the next T lines contains a single integer n_i (2 ≤ n_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of elegant numbers from 2 to n_i.

Example

Input

4
4
2
72
10


Output

2
1
61
6

Note

Here is the list of non-elegant numbers up to 10:

  * 4 = 2^2, GCD = 2; 
  * 8 = 2^3, GCD = 3; 
  * 9 = 3^2, GCD = 2. 



The rest have GCD = 1.
#include <bits/stdc++.h>
using namespace std;
const long long border = (long long)1e18;
vector<long long> p;
void init() {
  for (long long i = 2; i * i * i <= border; i++) {
    long long now = i * i;
    long long cnt = i;
    long long top = border / i / i;
    while (cnt <= top) {
      now *= i;
      cnt *= i;
      long long sqr = sqrt(now);
      if (sqr * sqr != now) {
        p.push_back(now);
      }
    }
  }
  sort(p.begin(), p.end());
  p.erase(unique(p.begin(), p.end()), p.end());
}
long long cal(long long r) {
  long long res = sqrt(r);
  res += upper_bound(p.begin(), p.end(), r) - p.begin();
  return res;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  init();
  int q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    cout << n - cal(n) << '\n';
  }
  return 0;
}
