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 int MAXN = 1e5;
const long long INF = 1e18;
const int MAXPW = 62;
const double EPS = 1e-9;
vector<long long> perfect;
inline bool IsPerfect(long long n) {
  long long x = sqrt(n);
  return x * x < n;
}
void precalc() {
  for (long long i = 2; i <= 1000000; ++i) {
    for (long long temp = i * i; temp <= INF / i; temp *= i)
      if (IsPerfect(temp * i)) perfect.push_back(temp * i);
  }
  sort(perfect.begin(), perfect.end());
  perfect.resize(unique(perfect.begin(), perfect.end()) - perfect.begin());
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  precalc();
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long sq = sqrt(n);
    if (sq * sq > n) sq--;
    long long pf =
        upper_bound(perfect.begin(), perfect.end(), n) - perfect.begin();
    cout << n - sq - pf << '\n';
  }
  return 0;
}
