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;
set<long long> inl;
bool isSquare(long long x) {
  long long a = (long long)(sqrt((double)x));
  return (a * a == x || (a + 1) * (a + 1) == x);
}
long long sqflr(long long x) {
  long long a = (long long)(sqrt((double)x)) + 2;
  while (a * a > x) a--;
  return a;
}
long long ex(long long a, int n) {
  long long res = 1;
  for (int i = 0; i < n; i++) {
    res *= a;
    if (res < 0 || res > 4e18 + 5) return -1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  for (int n = 3; n <= 64; n += 2) {
    for (int i = 2;; i++) {
      long long f = ex(i, n);
      if (f <= -1) break;
      if (!isSquare(i) && f > 0) inl.insert(f);
    }
  }
  int t;
  cin >> t;
  vector<long long> sorted;
  for (long long i : inl)
    if (i > 0) sorted.push_back(i);
  while (t--) {
    long long ni;
    cin >> ni;
    long long sq = sqflr(ni);
    auto it = upper_bound(sorted.begin(), sorted.end(), ni);
    long long inelegant = (long long)(distance(sorted.begin(), it)) + sq;
    cout << ni - inelegant << '\n';
  }
  return 0;
}
