#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
void solve() {
  map<int, int> mp;
  int n, res = 0;
  cin >> n;
  for (int i = 1; i <= sqrt(n); i++) {
    if (i <= 1000) {
      if (i * i <= n && !mp[i * i]) {
        res++;
        mp[i * i] = 1;
      }
      if (i * i * i <= n && !mp[i * i * i]) {
        res++;
        mp[i * i * i] = 1;
      }
    } else {
      if (i * i <= n && !mp[i * i]) {
        res++;
        mp[i * i] = 1;
      }
    }
  }
  cout << res << endl;
  return;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### 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 INF = 0x3f3f3f3f;
const int N = 1e6 + 10;
void solve() {
  map<int, int> mp;
  int n, res = 0;
  cin >> n;
  for (int i = 1; i <= sqrt(n); i++) {
    if (i <= 1000) {
      if (i * i <= n && !mp[i * i]) {
        res++;
        mp[i * i] = 1;
      }
      if (i * i * i <= n && !mp[i * i * i]) {
        res++;
        mp[i * i * i] = 1;
      }
    } else {
      if (i * i <= n && !mp[i * i]) {
        res++;
        mp[i * i] = 1;
      }
    }
  }
  cout << res << endl;
  return;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```