#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 10;
const long long INF = 2147483647;
const long long mod = 1e9 + 7;
const double eps = 1e-8;
const long long pi = 3.1415926;
long long n;
long long st[N];
void solve() {
  cin >> n;
  memset(st, 0, sizeof st);
  long long k = sqrt(n), q = 0;
  for (long long i = 1; i <= n / i; i++) {
    if (i * i * i <= n) st[q++] = i * i * i;
  }
  for (long long i = 0; i < q; i++) {
    long long t = sqrt(st[i]);
    if (t * t == st[i]) k--;
  }
  cout << k + q << '\n';
}
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return ~~(0 ^ 0 ^ 0);
}

### Prompt

Develop a solution in Cpp 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 N = 1e6 + 10;
const long long INF = 2147483647;
const long long mod = 1e9 + 7;
const double eps = 1e-8;
const long long pi = 3.1415926;
long long n;
long long st[N];
void solve() {
  cin >> n;
  memset(st, 0, sizeof st);
  long long k = sqrt(n), q = 0;
  for (long long i = 1; i <= n / i; i++) {
    if (i * i * i <= n) st[q++] = i * i * i;
  }
  for (long long i = 0; i < q; i++) {
    long long t = sqrt(st[i]);
    if (t * t == st[i]) k--;
  }
  cout << k + q << '\n';
}
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  cin >> T;
  while (T--) solve();
  return ~~(0 ^ 0 ^ 0);
}

```