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;
long long pw(long long x, long long p) {
  if (p == 0) return 1;
  long long t = pw(x, p / 2);
  if (p % 2 == 0)
    return (t * t);
  else
    return (((t * t)) * x);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, t;
  long long i, j;
  cin >> t;
  long long ans = 0;
  while (t--) {
    vector<long long> v(65, 0);
    cin >> n;
    ans = 0;
    for (i = 2; i <= 60; i++) {
      long long y;
      y = floor(((double)pow((double)n, (double)(1.0 / i))));
      while (pw(y, i) > 0 && pw(y, i) <= 1e18 && pw(y, i) <= n) y++;
      y--;
      y--;
      v[i] = y;
    }
    for (i = 60; i >= 2; i--)
      for (j = 2 * i; j <= 60; j += i) v[i] -= v[j];
    for (i = 2; i <= 60; i++) {
      ans += v[i];
    }
    cout << (n - 1) - ans << endl;
  }
  return 0;
}
