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;
vector<long long> v, vv;
void calc() {
  long long mx = 1e18;
  for (long long i = 2; i <= 1e6; i++) {
    for (long long u = i * i * i; u <= mx; u *= i) {
      v.push_back(u);
      if (u > mx / i) break;
    }
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < (int)v.size(); i++) {
    long long u = v[i];
    long long uu = sqrt(u);
    if (uu * uu == u) continue;
    vv.push_back(u);
  }
}
int main() {
  calc();
  int t;
  scanf("%d", &t);
  while (t--) {
    long long num, ans = 0;
    scanf("%lld", &num);
    ans += (long long)sqrt(num) - 1;
    long long all_a_to_b =
        (upper_bound(vv.begin(), vv.end(), num) - vv.begin());
    ans += all_a_to_b;
    ans = num - 1 - ans;
    printf("%lld\n", ans);
  }
}
