#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void chmax(T &x, T &y) {
  if (x < y) swap(x, y);
}
template <class T>
inline void chmin(T &x, T &y) {
  if (x > y) swap(x, y);
}
const long long INF = 0x3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
const double EPS = 10e-10;
const long long P2LIM = (long long)2e18;
const long long maxN = 1e9 + 10;
void Solution() {}
long long countSC(long long N) {
  long long res =
      (long long)sqrt(N) + (long long)cbrt(N) - (long long)(sqrt(cbrt(N)));
  return res;
}
int32_t main() {
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    cout << countSC(n) << '\n';
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <class T>
inline void chmax(T &x, T &y) {
  if (x < y) swap(x, y);
}
template <class T>
inline void chmin(T &x, T &y) {
  if (x > y) swap(x, y);
}
const long long INF = 0x3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
const double EPS = 10e-10;
const long long P2LIM = (long long)2e18;
const long long maxN = 1e9 + 10;
void Solution() {}
long long countSC(long long N) {
  long long res =
      (long long)sqrt(N) + (long long)cbrt(N) - (long long)(sqrt(cbrt(N)));
  return res;
}
int32_t main() {
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    cout << countSC(n) << '\n';
  }
  return 0;
}

```