#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    int a, b;
    for (int i = 1;; i++) {
      a = i;
      if (i * i > n) {
        break;
      }
    }
    for (int i = 1;; i++) {
      b = i;
      if (i * i * i > n) {
        break;
      }
    }
    a = a - 1;
    b = b - 1;
    int x, c;
    if (b >= 4) {
      for (int i = 1;; i++) {
        c = i;
        if (i * i > b) {
          break;
        }
      }
      c = c - 1;
      x = c - 1;
    }
    cout << a + b - x - 1 << endl;
    a = 0;
    b = 0;
    c = 0;
    x = 0;
  }
}

### Prompt

Your task is to create 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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    int a, b;
    for (int i = 1;; i++) {
      a = i;
      if (i * i > n) {
        break;
      }
    }
    for (int i = 1;; i++) {
      b = i;
      if (i * i * i > n) {
        break;
      }
    }
    a = a - 1;
    b = b - 1;
    int x, c;
    if (b >= 4) {
      for (int i = 1;; i++) {
        c = i;
        if (i * i > b) {
          break;
        }
      }
      c = c - 1;
      x = c - 1;
    }
    cout << a + b - x - 1 << endl;
    a = 0;
    b = 0;
    c = 0;
    x = 0;
  }
}

```