#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 5e5 + 7;
void hash0111() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int prevPowerOfTwo(long long int n) {
  if (n && (!n & (n - 1))) return n;
  return 1 << ((long long int)ceil(log2(n) - 1));
}
long long int nextPowerOfTwo(long long int n) {
  if (n && (!n & (n - 1))) return n;
  return 1 << ((long long int)ceil(log2(n)));
}
void solve() {
  long long int n;
  cin >> n;
  long long int res = 0;
  res += (long long int)sqrt(n);
  long long int x = (long long int)cbrt(n);
  res += x;
  res -= (long long int)sqrt(x);
  cout << res << '\n';
}
int main() {
  hash0111();
  long long int t;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long int MOD = 5e5 + 7;
void hash0111() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long int prevPowerOfTwo(long long int n) {
  if (n && (!n & (n - 1))) return n;
  return 1 << ((long long int)ceil(log2(n) - 1));
}
long long int nextPowerOfTwo(long long int n) {
  if (n && (!n & (n - 1))) return n;
  return 1 << ((long long int)ceil(log2(n)));
}
void solve() {
  long long int n;
  cin >> n;
  long long int res = 0;
  res += (long long int)sqrt(n);
  long long int x = (long long int)cbrt(n);
  res += x;
  res -= (long long int)sqrt(x);
  cout << res << '\n';
}
int main() {
  hash0111();
  long long int t;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```