#include <bits/stdc++.h>
using namespace std;
void init() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
void zerbooo() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
long long a[10000000] = {};
int main() {
  init();
  zerbooo();
  int n;
  cin >> n;
  while (n > 0) {
    long long t;
    cin >> t;
    vector<int> v1;
    for (int i = 0; i <= t; ++i) {
      if (i * i > t)
        break;
      else
        v1.push_back(i);
    }
    for (int i = 0; i * i * i <= t; ++i) {
      if (sqrt(i * i * i) - (int)sqrt(i * i * i) == 0)
        continue;
      else
        v1.push_back(i);
    }
    cout << v1.size() - 1 << "\n";
    --n;
  }
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

Polycarp likes squares and cubes of positive integers. Here is the beginning of the sequence of numbers he likes: 1, 4, 8, 9, ....

For a given number n, count the number of integers from 1 to n that Polycarp likes. In other words, find the number of such x that x is a square of a positive integer number or a cube of a positive integer number (or both a square and a cube simultaneously).

Input

The first line contains an integer t (1 ≤ t ≤ 20) — the number of test cases.

Then t lines contain the test cases, one per line. Each of the lines contains one integer n (1 ≤ n ≤ 10^9).

Output

For each test case, print the answer you are looking for — the number of integers from 1 to n that Polycarp likes.

Example

Input


6
10
1
25
1000000000
999999999
500000000


Output


4
1
6
32591
32590
23125

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void init() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
void zerbooo() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
long long a[10000000] = {};
int main() {
  init();
  zerbooo();
  int n;
  cin >> n;
  while (n > 0) {
    long long t;
    cin >> t;
    vector<int> v1;
    for (int i = 0; i <= t; ++i) {
      if (i * i > t)
        break;
      else
        v1.push_back(i);
    }
    for (int i = 0; i * i * i <= t; ++i) {
      if (sqrt(i * i * i) - (int)sqrt(i * i * i) == 0)
        continue;
      else
        v1.push_back(i);
    }
    cout << v1.size() - 1 << "\n";
    --n;
  }
  return 0;
}

```