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;
const long long Maxn = 1000000000000000000ll;
long long f[1004005];
bool check(long long x) {
  long long now = (long long)sqrt(x * 1.0);
  return now * now < x;
}
int main() {
  long long x;
  int len = 0;
  for (long long i = 2; i <= 1000000; i++) {
    x = i * i;
    while (x <= Maxn / i) {
      x *= i;
      if (check(x)) {
        f[++len] = x;
      }
    }
  }
  sort(f + 1, f + 1 + len);
  len = (int)(unique(f + 1, f + 1 + len) - f - 1);
  int t;
  scanf("%d", &t);
  long long n;
  while (t--) {
    cin >> n;
    x = upper_bound(f + 1, f + 1 + len, n) - f - 1;
    cout << n - x - (long long)sqrt(n * 1.0) << endl;
  }
  return 0;
}
