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;
long long a[(30000000 + 5000)], tot = 0, T, n;
int squ(long long N) {
  long long Rt = sqrt(N);
  if (Rt * Rt == N) return 1;
  return 0;
}
int main() {
  for (long long i = 2; i <= 1000000; ++i) {
    long long tmp = i * i;
    while (tmp <= ((long long)(1e18)) / i) {
      tmp *= i;
      if (!squ(tmp)) a[++tot] = tmp;
    }
  }
  sort(a + 1, a + 1 + tot);
  tot = (long long)(unique(a + 1, a + 1 + tot) - a - 1);
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld", &n);
    long long pos = upper_bound(a + 1, a + 1 + tot, n) - a - 1;
    printf("%lld\n", n - pos - (long long)sqrt(n));
  }
  return 0;
}
