#include <bits/stdc++.h>
using namespace std;
using cd = complex<double>;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const long long N = 1e5 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  set<long long> s;
  vector<long long> p;
  for (long long i = 1; i <= 100000; i++) {
    if (i * i <= 1e9) s.insert(i * i);
    if (i * i * i <= 1e9) s.insert(i * i * i);
  }
  for (auto i : s) {
    p.push_back(i);
  }
  while (q--) {
    long long n;
    cin >> n;
    long long l = upper_bound(p.begin(), p.end(), n) - p.begin();
    cout << l << endl;
  }
}

### Prompt

Your task is to create 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;
using cd = complex<double>;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const long long N = 1e5 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  set<long long> s;
  vector<long long> p;
  for (long long i = 1; i <= 100000; i++) {
    if (i * i <= 1e9) s.insert(i * i);
    if (i * i * i <= 1e9) s.insert(i * i * i);
  }
  for (auto i : s) {
    p.push_back(i);
  }
  while (q--) {
    long long n;
    cin >> n;
    long long l = upper_bound(p.begin(), p.end(), n) - p.begin();
    cout << l << endl;
  }
}

```