#include <bits/stdc++.h>
using namespace std;
int T, mu[99];
long long n, ans;
long long gen(long long n, long long k) {
  long long t = powl(n, 1. / k) - 0.5;
  return t + (powl(t + 1, k) - 0.5 <= n);
}
int main() {
  for (int i = 2; i < 60; ++i) {
    mu[i] = -1 - mu[i];
    for (int j = i + i; j < 60; j += i) mu[j] += mu[i];
  }
  for (cin >> T; T--;) {
    cin >> n;
    ans = n - 1;
    for (int i = 2; i < 60; ++i) ans += mu[i] * (gen(n, i) - 1);
    cout << ans << endl;
  }
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int T, mu[99];
long long n, ans;
long long gen(long long n, long long k) {
  long long t = powl(n, 1. / k) - 0.5;
  return t + (powl(t + 1, k) - 0.5 <= n);
}
int main() {
  for (int i = 2; i < 60; ++i) {
    mu[i] = -1 - mu[i];
    for (int j = i + i; j < 60; j += i) mu[j] += mu[i];
  }
  for (cin >> T; T--;) {
    cin >> n;
    ans = n - 1;
    for (int i = 2; i < 60; ++i) ans += mu[i] * (gen(n, i) - 1);
    cout << ans << endl;
  }
}

```