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 n;
vector<long long> v;
int sqrt(long long x) {
  int low = 1;
  int high = 1e9;
  while (low < high) {
    int mid = low + (high - low + 1) / 2;
    if (1LL * mid * mid <= x) {
      low = mid;
    } else {
      high = mid - 1;
    }
  }
  return low;
}
int main() {
  int t;
  cin >> t;
  for (int i = 2; i <= 1000000; i++) {
    for (long long j = 1LL * i * i * i;; j *= i) {
      long long temp = sqrt(j);
      if (temp * temp != j) {
        v.push_back(j);
      }
      if (j > (LLONG_MAX / i)) {
        break;
      }
    }
  }
  sort(v.begin(), v.end());
  unique(v.begin(), v.end());
  while (t--) {
    cin >> n;
    long long temp = sqrt(n) - 1;
    int temp2 = upper_bound(v.begin(), v.end(), n) - v.begin();
    long long ans = n - temp - temp2 - 1;
    cout << ans << "\n";
  }
  return 0;
}
