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;
bool isprime[64];
long long mobius[64];
long long root(long long n, long long i) {
  long double n_eps = double(n) + 0.001;
  long double n_fl = n;
  long long res = pow(n_fl, 1.0 / float(i));
  while (pow(res + 1, i) < n_eps) {
    res++;
  }
  while (pow(res, i) > n_fl) {
    res--;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  long long n, r, f;
  int t, pc;
  cin >> t;
  for (int i = 2; i < 64; i++) {
    isprime[i] = true;
    mobius[i] = 1;
    pc = 0;
    for (int j = 2; j < i; j++) {
      if (i % j == 0) {
        isprime[i] = false;
        if (isprime[j]) {
          pc++;
        }
        if (isprime[j] && (i % (j * j) == 0)) {
          mobius[i] = 0;
        }
      }
    }
    if (!isprime[i] && (mobius[i] == 1) && (pc % 2 == 0)) {
      mobius[i] = -1;
    }
  }
  while (t--) {
    cin >> n;
    r = n - 1;
    for (int i = 2; i < 64; i++) {
      f = root(n, i);
      r -= mobius[i] * (f - 1);
      if (f == 1) {
        break;
      }
    }
    cout << r << endl;
  }
  return 0;
}
