#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
int mi2(int x1) {
  long long sum = 0;
  int cnt = 1;
  while (sum <= x1) {
    cnt++;
    sum = cnt * cnt;
  }
  return cnt - 1;
}
int mi3(int x1) {
  long long sum = 0;
  int cnt = 1;
  while (sum <= x1) {
    cnt++;
    sum = cnt * cnt * cnt;
  }
  return cnt - 1;
}
int mi6(int x1) {
  long long sum = 0;
  int cnt = 1;
  while (sum <= x1) {
    cnt++;
    sum = cnt * cnt * cnt * cnt * cnt * cnt;
  }
  return cnt - 1;
}
int main(void) {
  int t;
  int n;
  int put;
  cin >> t;
  for (int i = 1; i <= (t); i++) {
    cin >> n;
    int x1, x2, x3;
    x1 = int(pow(n + 0.5, 1.0 / 2));
    x2 = int(pow(n + 0.5, 1.0 / 3));
    x3 = int(pow(n + 0.5, 1.0 / 6));
    put = mi2(n) + mi3(n) - mi6(n);
    cout << x1 + x2 - x3 << "\n";
  }
}

### Prompt

Please create a solution in Cpp 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;
const double pi = acos(-1.0);
int mi2(int x1) {
  long long sum = 0;
  int cnt = 1;
  while (sum <= x1) {
    cnt++;
    sum = cnt * cnt;
  }
  return cnt - 1;
}
int mi3(int x1) {
  long long sum = 0;
  int cnt = 1;
  while (sum <= x1) {
    cnt++;
    sum = cnt * cnt * cnt;
  }
  return cnt - 1;
}
int mi6(int x1) {
  long long sum = 0;
  int cnt = 1;
  while (sum <= x1) {
    cnt++;
    sum = cnt * cnt * cnt * cnt * cnt * cnt;
  }
  return cnt - 1;
}
int main(void) {
  int t;
  int n;
  int put;
  cin >> t;
  for (int i = 1; i <= (t); i++) {
    cin >> n;
    int x1, x2, x3;
    x1 = int(pow(n + 0.5, 1.0 / 2));
    x2 = int(pow(n + 0.5, 1.0 / 3));
    x3 = int(pow(n + 0.5, 1.0 / 6));
    put = mi2(n) + mi3(n) - mi6(n);
    cout << x1 + x2 - x3 << "\n";
  }
}

```