#include <bits/stdc++.h>
using namespace std;
using LL = long long;
constexpr int N = 1e5;
constexpr int INF = 0x3f3f3f3f;
vector<LL> seq;
unordered_set<LL> st;
inline void init() {
  for (LL i = 1; i <= N; ++i) {
    if (!st.count(i * i)) {
      st.insert(i * i);
      seq.push_back(i * i);
    }
    if (!st.count(i * i * i)) {
      st.insert(i * i * i);
      seq.push_back(i * i * i);
    }
  }
  sort(seq.begin(), seq.end());
}
void solve() {
  int n;
  cin >> n;
  int l = 0, r = seq.size() - 1;
  while (l < r) {
    int mid = l + r + 1 >> 1;
    if (seq[mid] <= n)
      l = mid;
    else
      r = mid - 1;
  }
  cout << l + 1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

### 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;
using LL = long long;
constexpr int N = 1e5;
constexpr int INF = 0x3f3f3f3f;
vector<LL> seq;
unordered_set<LL> st;
inline void init() {
  for (LL i = 1; i <= N; ++i) {
    if (!st.count(i * i)) {
      st.insert(i * i);
      seq.push_back(i * i);
    }
    if (!st.count(i * i * i)) {
      st.insert(i * i * i);
      seq.push_back(i * i * i);
    }
  }
  sort(seq.begin(), seq.end());
}
void solve() {
  int n;
  cin >> n;
  int l = 0, r = seq.size() - 1;
  while (l < r) {
    int mid = l + r + 1 >> 1;
    if (seq[mid] <= n)
      l = mid;
    else
      r = mid - 1;
  }
  cout << l + 1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```