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>
#pragma gcc optimize("Ofast")
using namespace std;
vector<long long> v;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (long long i = 2; i <= 1e6; ++i) {
    long long res = i;
    while (res <= 1e18 && log10(res) + log10(i) <= 18) {
      res *= i;
      long long s = sqrt(res);
      if (s * s != res) v.emplace_back(res);
    }
  }
  sort((v).begin(), (v).end());
  v.erase(unique((v).begin(), (v).end()), v.end());
  long long t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    cout << (n - 1) -
                ((upper_bound((v).begin(), (v).end(), n) - v.begin() - 1) +
                 (long long)sqrt(n))
         << '\n';
  }
}
