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;
vector<long long> all;
inline bool even(long long x) {
  long long y = sqrtl(x);
  return y * y == x;
}
inline void solve() {
  long long n;
  cin >> n;
  long long y = sqrtl(n);
  cout << n - y - (upper_bound(all.begin(), all.end(), n) - all.begin())
       << '\n';
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  const long long LIM = 1e18 + 1;
  for (long long i = 2; i <= 1e6; i++) {
    long long cur = i * i * i;
    while (1) {
      if (!even(cur)) all.push_back(cur);
      if (LIM / i < cur) break;
      cur *= i;
    }
  }
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}
