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;
  using vll = vector<long long>;
  using vi = vector<long long>;
  using vpl = vector<pair<long long, long long>>;
  using pll = pair<long long, long long>;
  long long mod = 1000000007;
  long long mod2 = 998244353;
  const long long MX = 2e5 + 5;
  long long ipow(long long base, long long k, long long M) {
    base %= M;
    k = k % (M - 1);
    long long total = 1;
    while (k) {
      if (k & 1) {
        total = (total * base) % M;
      }
      base = (base * base) % M;
      k >>= 1;
    }
    return total;
  }
  long long precomp[100];
  long long modu(long long a, long long b) {
    long long c = a % b;
    return (c < 0) ? c + b : c;
  }
  void solve() {
    long long v, e, a, b;
    long long n, bal;
    long long count = 0, x, y;
    cin >> n;
    a = floor(sqrt(n));
    b = floor(cbrt(n));
    long long ans = a + b;
    for (long long i = 0; i < 101; i++) {
      if (n < precomp[i]) {
        ans -= i;
        break;
      }
    }
    cout << ans << endl;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long tc = 1;
    long long cnt = 1;
    cin >> tc;
    for (long long i = 1; i < 101; i++) {
      precomp[i - 1] = ipow(i, 6, 1000000000000000000);
    }
    while (tc--) {
      solve();
    }
  }
  