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 MAXN = 300100;
vector<long long> vec;
void precalc() {
  for (long long i = 2; i < 1000000; i++) {
    long long x = 1ll * i * i;
    while (x + 0.0 < (2e18 + 100) / i) {
      x *= 1ll * i;
      if ((long long)sqrt(x) * (long long)(sqrt(x)) != x) vec.push_back(x);
    }
  }
  sort((vec).begin(), (vec).end());
  unique((vec).begin(), (vec).end());
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long i = 0;
  precalc();
  long long t;
  cin >> t;
  while (t--) {
    long long x;
    cin >> x;
    cout << x - (upper_bound((vec).begin(), (vec).end(), x) - vec.begin() +
                 (long long)sqrt(x))
         << '\n';
  }
}
