#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
int m;
int c[N];
int cc[N];
int main() {
  cin >> n;
  while (n--) {
    int a;
    cin >> a;
    int sum = 0;
    sum = sqrt(a) + (1e-9);
    sum += pow(a, (1.0) / 3) + (1e-9);
    for (int i = 1; i <= pow(a, (1.0) / 3) + (1e-9); i++) {
      if (abs(sqrt(i) - (int)(sqrt(i) + 0.5)) <= 1e-9 &&
          abs(sqrt(i) - (int)(sqrt(i) + 0.5)) >= -1e-9)
        sum--;
    }
    cout << sum << "\n";
  }
}

### Prompt

In CPP, your task is to solve 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 N = 2e5 + 5;
int n;
int m;
int c[N];
int cc[N];
int main() {
  cin >> n;
  while (n--) {
    int a;
    cin >> a;
    int sum = 0;
    sum = sqrt(a) + (1e-9);
    sum += pow(a, (1.0) / 3) + (1e-9);
    for (int i = 1; i <= pow(a, (1.0) / 3) + (1e-9); i++) {
      if (abs(sqrt(i) - (int)(sqrt(i) + 0.5)) <= 1e-9 &&
          abs(sqrt(i) - (int)(sqrt(i) + 0.5)) >= -1e-9)
        sum--;
    }
    cout << sum << "\n";
  }
}

```