#include <bits/stdc++.h>
using namespace std;
const long long n = 1e9;
vector<long long> p;
map<long long, long long> q;
set<long long> r;
long long binary(long long a) {
  long long j = 1, count = 0;
  for (long long i = 1; i <= a; i++) {
    long long p = 6, res = 1, j = i;
    while (p > 0) {
      if (p & 1) res = res * j;
      j = j * j;
      p >>= 1;
    }
    if (res <= a) {
      count++;
    } else {
      return count;
      break;
    }
  }
}
void solve() {
  long long a;
  cin >> a;
  long long b = sqrt(a);
  long long c = cbrt(a);
  if (a != 1) {
    long long x = binary(a);
    cout << b + c - x << endl;
  } else
    cout << "1" << endl;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your challenge is to write 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;
const long long n = 1e9;
vector<long long> p;
map<long long, long long> q;
set<long long> r;
long long binary(long long a) {
  long long j = 1, count = 0;
  for (long long i = 1; i <= a; i++) {
    long long p = 6, res = 1, j = i;
    while (p > 0) {
      if (p & 1) res = res * j;
      j = j * j;
      p >>= 1;
    }
    if (res <= a) {
      count++;
    } else {
      return count;
      break;
    }
  }
}
void solve() {
  long long a;
  cin >> a;
  long long b = sqrt(a);
  long long c = cbrt(a);
  if (a != 1) {
    long long x = binary(a);
    cout << b + c - x << endl;
  } else
    cout << "1" << endl;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```