#include <bits/stdc++.h>
using namespace std;
void _print(long long int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(double t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <typename T>
istream &operator>>(istream &is, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a << " ";
  }
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p) {
  os << p.first << " " << p.second << '\n';
  return os;
}
template <typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, vector<pair<T1, T2>> &arr) {
  typename vector<pair<T1, T2>>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a;
  }
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, vector<pair<T1, T2>> &arr) {
  typename vector<pair<T1, T2>>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<bool> isprime(2e5 + 10, true);
vector<long long int> lp(2e5 + 10), hp(2e5 + 10);
void sieve() {
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < 2e5 + 10; i++) {
    if (isprime[i] == true) {
      for (long long int j = 2 * i; j < 2e5 + 10; j += i) {
        isprime[j] = false;
        hp[j] = i;
        if (lp[j] == 0) lp[j] = i;
      }
    }
  }
}
long long int add(long long int a, long long int b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long int sub(long long int a, long long int b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long int mul(long long int a, long long int b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long int mpow(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
void solve() {
  long long int n;
  cin >> n;
  long long int cnt = 1;
  long long int i = 2;
  long long int ans = 0;
  while (n >= i * i) {
    i++;
    cnt++;
  }
  long long int j = 2;
  while (n >= j * j * j) {
    j++;
    cnt++;
  }
  long long int k = 2;
  while (n >= pow(k, 6)) {
    k++;
    ans++;
  }
  cout << cnt - ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  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;
void _print(long long int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(double t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <typename T>
istream &operator>>(istream &is, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a << " ";
  }
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p) {
  os << p.first << " " << p.second << '\n';
  return os;
}
template <typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, vector<pair<T1, T2>> &arr) {
  typename vector<pair<T1, T2>>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a;
  }
  return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, vector<pair<T1, T2>> &arr) {
  typename vector<pair<T1, T2>>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<bool> isprime(2e5 + 10, true);
vector<long long int> lp(2e5 + 10), hp(2e5 + 10);
void sieve() {
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < 2e5 + 10; i++) {
    if (isprime[i] == true) {
      for (long long int j = 2 * i; j < 2e5 + 10; j += i) {
        isprime[j] = false;
        hp[j] = i;
        if (lp[j] == 0) lp[j] = i;
      }
    }
  }
}
long long int add(long long int a, long long int b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long int sub(long long int a, long long int b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long int mul(long long int a, long long int b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long int mpow(long long int x, long long int y) {
  long long int res = 1;
  x = x % 1000000007;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
void solve() {
  long long int n;
  cin >> n;
  long long int cnt = 1;
  long long int i = 2;
  long long int ans = 0;
  while (n >= i * i) {
    i++;
    cnt++;
  }
  long long int j = 2;
  while (n >= j * j * j) {
    j++;
    cnt++;
  }
  long long int k = 2;
  while (n >= pow(k, 6)) {
    k++;
    ans++;
  }
  cout << cnt - ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```