#include <bits/stdc++.h>
using namespace std;
void sol() {
  long long n;
  cin >> n;
  long long count = 1;
  long long mxn = 31623;
  for (long long i = 2; i < mxn; i++) {
    if ((i * i) <= n) count++;
    long long temp = i * i * i;
    long long temp2 = sqrt(temp);
    if (temp <= n && temp != (temp2 * temp2)) count++;
  }
  cout << count << "\n";
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    sol();
  }
}

### Prompt

Please formulate 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 sol() {
  long long n;
  cin >> n;
  long long count = 1;
  long long mxn = 31623;
  for (long long i = 2; i < mxn; i++) {
    if ((i * i) <= n) count++;
    long long temp = i * i * i;
    long long temp2 = sqrt(temp);
    if (temp <= n && temp != (temp2 * temp2)) count++;
  }
  cout << count << "\n";
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    sol();
  }
}

```