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> contains;
const long long MAX = 1e18 + 1e5;
bool square(long long x) {
  long long y = sqrt(x);
  if (y * y == x) return 1;
  return 0;
}
void Init() {
  for (int base = (2); base <= (1e6); ++base) {
    long long now = 1ll * base * base;
    while (1.0 * MAX / now > base) {
      now *= base;
      if (!square(now)) contains.push_back(now);
    }
  }
  sort(contains.begin(), contains.end());
  contains.resize(unique(contains.begin(), contains.end()) - contains.begin());
}
void Query() {
  long long n;
  cin >> n;
  long long ans = n - (long long)sqrt(n);
  ans -= upper_bound(contains.begin(), contains.end(), n) - contains.begin();
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Init();
  int test;
  cin >> test;
  while (test--) Query();
}
