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> st;
const long long mx = 1e18;
std::vector<long long> v;
void pre() {
  for (long long i = 2; i <= 1000006; i++) {
    for (long long j = i * i * i; j <= mx; j *= i) {
      long long g = sqrt(j);
      if (g * g != j) {
        st.insert(j);
      }
      if (j > mx / i) break;
    }
  }
  v.assign(st.begin(), st.end());
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  long long T;
  cin >> T;
  pre();
  while (T--) {
    long long n;
    cin >> n;
    long long ans = upper_bound(v.begin(), v.end(), n) -
                    upper_bound(v.begin(), v.end(), 2 - 1);
    ans += sqrt(n) - sqrt(2 - 1);
    ans = n - 2 + 1 - ans;
    cout << ans << "\n";
  }
  return 0;
}
