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 double eps = 1e-10;
const double Pi = 3.1415926535897932;
const int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};
const int maxn = 2e6;
long long w[maxn];
int pn = 0;
const long long N = 1e18;
int main() {
  for (long long i = 2; i <= 1e6; i++) {
    long long t = i * i;
    while (t <= N / i) {
      t *= i;
      long long tmp = sqrt(t);
      if (tmp * tmp < t) w[++pn] = t;
    }
  }
  sort(w + 1, w + 1 + pn);
  pn = unique(w + 1, w + 1 + pn) - w - 1;
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n;
    scanf("%lld", &n);
    long long ans = n - ((long long)sqrt(n));
    ans -= upper_bound(w + 1, w + 1 + pn, n) - w - 1;
    printf("%lld\n", ans);
  }
  return 0;
}
