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;
int mu[66];
void init() {
  mu[1] = 1;
  for (int i = 1; i <= 59; i++)
    for (int j = i * 2; j <= 59; j += i) mu[j] -= mu[i];
}
void solve() {
  long long n, ans;
  scanf("%lld", &n);
  ans = n - 1;
  for (int i = 2; i <= 59; i++) {
    if (!mu[i]) continue;
    long long limit = (long long)pow(n, 1.0 / i);
    for (;;) {
      limit++;
      long long tmp = n;
      for (int j = 1; j <= i; j++) tmp /= limit;
      if (!tmp) {
        limit--;
        break;
      }
    }
    ans += (limit - 1) * mu[i];
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  int T;
  scanf("%d", &T);
  while (T--) solve();
}
