#include <bits/stdc++.h>
const long long MXN = 1e5 + 7;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
using namespace std;
long long t, a[MXN], b[MXN], sum, id;
bool used[MXN];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    int n, id = 0;
    cin >> n;
    set<int> st;
    for (int i = 1; i <= sqrt(n); i++) {
      if (st.find(i * i) == st.end()) {
        id++;
        st.insert(i * i);
      }
    }
    for (int i = 1; i <= cbrt(n); i++) {
      if (st.find(i * i * i) == st.end()) {
        id++;
        st.insert(i * i * i);
      }
    }
    cout << id << '\n';
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
const long long MXN = 1e5 + 7;
const long long mod = 1e9 + 7;
const long long inf = 1e9;
using namespace std;
long long t, a[MXN], b[MXN], sum, id;
bool used[MXN];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    int n, id = 0;
    cin >> n;
    set<int> st;
    for (int i = 1; i <= sqrt(n); i++) {
      if (st.find(i * i) == st.end()) {
        id++;
        st.insert(i * i);
      }
    }
    for (int i = 1; i <= cbrt(n); i++) {
      if (st.find(i * i * i) == st.end()) {
        id++;
        st.insert(i * i * i);
      }
    }
    cout << id << '\n';
  }
}

```