Consider some positive integer x. Its prime factorization will be of form x = 2^{k_1} ⋅ 3^{k_2} ⋅ 5^{k_3} ⋅ ...

Let's call x elegant if the greatest common divisor of the sequence k_1, k_2, ... is equal to 1. For example, numbers 5 = 5^1, 12 = 2^2 ⋅ 3, 72 = 2^3 ⋅ 3^2 are elegant and numbers 8 = 2^3 (GCD = 3), 2500 = 2^2 ⋅ 5^4 (GCD = 2) are not.

Count the number of elegant integers from 2 to n.

Each testcase contains several values of n, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (1 ≤ T ≤ 10^5) — the number of values of n in the testcase.

Each of the next T lines contains a single integer n_i (2 ≤ n_i ≤ 10^{18}).

Output

Print T lines — the i-th line should contain the number of elegant numbers from 2 to n_i.

Example

Input

4
4
2
72
10


Output

2
1
61
6

Note

Here is the list of non-elegant numbers up to 10:

  * 4 = 2^2, GCD = 2; 
  * 8 = 2^3, GCD = 3; 
  * 9 = 3^2, GCD = 2. 



The rest have GCD = 1.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream &operator<<(ostream &os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
const int N = 101;
int mu[N], pr[N];
unsigned long long powr(unsigned long long n, unsigned long long p) {
  unsigned long long ans = 1;
  while (p) {
    if (p & 1) {
      ans = (ans * n);
    }
    n = (n * n);
    p >>= 1;
  }
  return ans;
}
void pew() {
  for (int i = 1; i < (101); ++i) pr[i] = i;
  for (int i = 2; i < (11); ++i) {
    if (pr[i] == i) {
      for (int j = i * i; j <= 100; j += i) {
        pr[j] = i;
      }
    }
  }
  mu[1] = 1;
  for (int i = 2; i < (101); ++i) {
    if ((i / pr[i]) % pr[i]) {
      mu[i] = -mu[i / pr[i]];
    }
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  pew();
  int tt;
  for (cin >> tt; tt--;) {
    unsigned long long n;
    cin >> n;
    unsigned long long ans = n - 1;
    for (int i = 2; i < (61); ++i)
      if (mu[i]) {
        unsigned long long x = floor(pow(n, 1.0L / i));
        if (pow(x + 1, i) <= 1e18 and powr(x + 1, i) <= n) ++x;
        ans += mu[i] * (x - 1);
      }
    cout << ans << '\n';
  }
  return 0;
}
