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 int mt(long long int a, long long int b) {
  if (a > 1e+18 / b)
    return -1;
  else
    return a * b;
}
long long int pwr(long long int x, long long int n) {
  long long int res = 1ll;
  while (n > 0) {
    if (n % 2 == 1) res *= x;
    x = (x * x);
    n /= 2;
  }
  return res;
}
bool check(long long int y) {
  long long int x = sqrt(y);
  if (x * x == y)
    return true;
  else
    return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  long long int n;
  long long int y;
  vector<long long int> v;
  map<long long int, long long int> map1;
  for (long long int i = 2; i * i * i <= 1e+18; i++) {
    y = i * i;
    while (1) {
      if (check(y) == false && map1.find(y) == map1.end()) {
        v.push_back(y), map1[y] = 1;
      }
      if (y > 1e+18 / i)
        break;
      else
        y *= i;
    }
  }
  sort(v.begin(), v.end());
  while (t--) {
    cin >> n;
    long long int z = sqrt(n);
    long long int ans = z + (upper_bound(v.begin(), v.end(), n) - v.begin());
    cout << (n - ans) << "\n";
  }
  return 0;
}
