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;
  long long dp[100000];
  long long mod = 998244353;
  void sieve() {
    memset(dp, true, sizeof(dp));
    for (long long i = 2; i < 100000; i++) {
      if (!dp[i]) continue;
      for (long long j = i * i; j < 10000; j += i) {
        dp[j] = false;
      }
    }
  }
  void reverse(long long i, long long j, long long a[]) {
    while (i < j) {
      swap(a[i], a[j]);
      i++, j--;
    }
  }
  bool ispallindrome(string s) {
    long long n = s.length();
    long long half = ceil(n / 2);
    for (long long i = 0; i < half; i++) {
      if (s[i] != s[n - i - 1]) return false;
    }
    return true;
  }
  long long power(long long n, long long p) {
    if (p == 0) return 1;
    if (n == 1) return 1;
    long long x = power(n, p / 2);
    if (p % 2 == 0) return (x * x) % mod;
    return ((n * x) % mod) * x % mod;
  }
  bool comp(pair<long long, long long> a, pair<long long, long long> b) {
    if (a.first == b.first) return a.second < b.second;
    return (a.first < b.first);
  }
  struct cmp1 {
    bool operator()(pair<long long, long long> a, pair<long long, long long> b) {
      if (a.first == b.first) return a.second < b.second;
      return a.first > b.first;
    }
  };
  void solve() {
    long long n;
    cin >> n;
    long long n1 = sqrt(n);
    long long n2 = cbrt(n);
    long long n3 = sqrt(n2);
    long long ans = n1 + n2 - n3;
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    long long t;
    cin >> t;
    long long t1 = t;
    while (t--) {
      solve();
    }
    return 0;
  }
  