#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<long long int> v;
  map<long long int, long long int> mp;
  for (long long int i = 1; i <= 100000; i++) {
    if (mp[i * i] == 0) {
      v.push_back((i * i));
      mp[i * i] = 1;
    }
    if (mp[i * i * i] == 0) {
      v.push_back((i * i * i));
      mp[i * i * i] = 1;
    }
  }
  sort(v.begin(), v.end());
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int u = upper_bound(v.begin(), v.end(), n) - v.begin();
    cout << u << "\n";
  }
}

### Prompt

Please formulate a CPP solution 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;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<long long int> v;
  map<long long int, long long int> mp;
  for (long long int i = 1; i <= 100000; i++) {
    if (mp[i * i] == 0) {
      v.push_back((i * i));
      mp[i * i] = 1;
    }
    if (mp[i * i * i] == 0) {
      v.push_back((i * i * i));
      mp[i * i * i] = 1;
    }
  }
  sort(v.begin(), v.end());
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int u = upper_bound(v.begin(), v.end(), n) - v.begin();
    cout << u << "\n";
  }
}

```