description: 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
solution: #include <bits/stdc++.h>
  using namespace std;
  const long long int mod = 1000000007;
  const long long int N = 1e9 + 2;
  long long int gcd(long long int a, long long int b) {
    if (b == 0)
      return a;
    else
      return gcd(b, a % b);
  }
  long long int bin_exp(long long int a, long long int b) {
    long long int res = 1;
    if (b == 0) return 1;
    while (b) {
      if (b % 2 == 0) {
        a = (a % mod) * (a % mod);
        b /= 2;
      } else {
        res = (res % mod) * (a % mod);
        b--;
      }
    }
    return res;
  }
  long long int mod_inv(long long int a, long long int b) {
    return bin_exp(a, b - 2) % b;
  }
  long long int fact[100];
  void cal() {
    fact[0] = 1;
    for (long long int i = 1; i < N; i++) {
      fact[i] = (fact[i - 1] % mod * i % mod) % mod;
    }
  }
  void solve() {
    long long int n;
    cin >> n;
    long long int x = int32_t(sqrt(n)) + int32_t(cbrt(n));
    cout << x - int32_t(sqrt(cbrt(n))) << endl;
  }
  int32_t main() {
    long long int t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  