#include <bits/stdc++.h>
using namespace std;
long long mx = 1e7;
vector<long long> primelist;
vector<bool> prime(mx + 5, true);
void sieve() {
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= mx; i++) {
    if (prime[i] == true) {
      primelist.push_back(i);
      for (long long j = i * i; j <= mx; j += i) {
        prime[j] = false;
      }
    }
  }
}
void solve() {
  long long i, j, n, k, c = 0, cube = 0;
  cin >> n;
  c = (long long)sqrt(n);
  k = (long long)cbrt(n);
  cout << c + k - (long long)sqrt(k) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t, i = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
long long mx = 1e7;
vector<long long> primelist;
vector<bool> prime(mx + 5, true);
void sieve() {
  prime[0] = prime[1] = false;
  for (long long i = 2; i <= mx; i++) {
    if (prime[i] == true) {
      primelist.push_back(i);
      for (long long j = i * i; j <= mx; j += i) {
        prime[j] = false;
      }
    }
  }
}
void solve() {
  long long i, j, n, k, c = 0, cube = 0;
  cin >> n;
  c = (long long)sqrt(n);
  k = (long long)cbrt(n);
  cout << c + k - (long long)sqrt(k) << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t, i = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```