#include <bits/stdc++.h>
using namespace std;
const int dx[9] = {-1, 0, 1, 0, -1, -1, 1, 1, 0};
const int dy[9] = {0, 1, 0, -1, -1, 1, -1, 1, 0};
long long n;
void solve() {
  long long a, b, c;
  {
    long long left = 0, right = 1e5;
    while (left < right) {
      long long mid = left + right + 1 >> 1;
      long long y = mid * mid;
      if (y <= n)
        left = mid;
      else
        right = mid - 1;
    }
    a = left;
  }
  {
    long long left = 0, right = 1e5;
    while (left < right) {
      long long mid = left + right + 1 >> 1;
      long long y = mid * mid * mid;
      if (y <= n)
        left = mid;
      else
        right = mid - 1;
    }
    b = left;
  }
  {
    long long left = 0, right = 1e3;
    while (left < right) {
      long long mid = left + right + 1 >> 1;
      long long y = mid * mid * mid * mid * mid * mid;
      if (y <= n)
        left = mid;
      else
        right = mid - 1;
    }
    c = left;
  }
  cout << a + b - c << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int dx[9] = {-1, 0, 1, 0, -1, -1, 1, 1, 0};
const int dy[9] = {0, 1, 0, -1, -1, 1, -1, 1, 0};
long long n;
void solve() {
  long long a, b, c;
  {
    long long left = 0, right = 1e5;
    while (left < right) {
      long long mid = left + right + 1 >> 1;
      long long y = mid * mid;
      if (y <= n)
        left = mid;
      else
        right = mid - 1;
    }
    a = left;
  }
  {
    long long left = 0, right = 1e5;
    while (left < right) {
      long long mid = left + right + 1 >> 1;
      long long y = mid * mid * mid;
      if (y <= n)
        left = mid;
      else
        right = mid - 1;
    }
    b = left;
  }
  {
    long long left = 0, right = 1e3;
    while (left < right) {
      long long mid = left + right + 1 >> 1;
      long long y = mid * mid * mid * mid * mid * mid;
      if (y <= n)
        left = mid;
      else
        right = mid - 1;
    }
    c = left;
  }
  cout << a + b - c << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    solve();
  }
  return 0;
}

```