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;
long long sq(long long t) {
  int d = 31;
  long long ans = 0;
  while (d--)
    if (ans + (1ll << d) <= 1000000000) {
      if ((ans + (1ll << d)) * (ans + (1ll << d)) <= t) ans += (1ll << d);
    }
  return ans;
}
void init() {
  for (int i = 2; i <= 1000000; ++i) {
    long long t = i * 1ll * i * i;
    while (t <= 1000000000000000000ll) {
      if (sq(t) * sq(t) != t) v.push_back(t);
      if (t * 1.0 * i <= 1000000000000000000ll)
        t *= i;
      else
        break;
    }
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
}
long long solve(long long t) {
  return sq(t) + (long long)(upper_bound(v.begin(), v.end(), t) - v.begin());
}
int main() {
  init();
  int q;
  scanf("%d", &q);
  while (q--) {
    long long n;
    scanf("%lld", &n);
    printf("%lld\n", n - solve(n));
  }
  return 0;
}
