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 double exp(long double a, long long int b) {
  long double ans = 1, x = a;
  while (b) {
    if (b % 2) ans *= x;
    x *= x;
    b = b >> 1;
  }
  return ans;
}
long long int power(long double n, long long int b) {
  long long int lo = 1, hi = sqrt(n);
  if (b > 18) hi = min(hi, 10LL);
  long long int ans = 0;
  while (lo <= hi) {
    long long int mid = (lo + hi) >> 1;
    if (exp(mid, b) > n) {
      hi = mid - 1;
    } else {
      ans = max(ans, mid);
      lo = mid + 1;
    }
  }
  return ans;
}
int prime[] = {2,  3,  5,  7,  11, 13, 17, 19, 23,
               29, 31, 37, 41, 43, 47, 53, 59, 61};
int k = 18;
int main() {
  ios_base::sync_with_stdio(false);
  vector<vector<int> > v(4);
  for (int i = 0; i < k; i++) v[1].push_back(prime[i]);
  for (int i = 0; i < k; i++)
    for (int j = i + 1; j < k; j++) {
      if (prime[i] * prime[j] < 61) {
        v[2].push_back(prime[i] * prime[j]);
      }
    }
  v[3].push_back(30);
  v[3].push_back(42);
  int t;
  cin >> t;
  while (t--) {
    long double n;
    cin >> n;
    long long int ans = n - 1;
    for (int i = 0; i < v[1].size(); i++) {
      long double h = power(n, v[1][i]);
      ans -= h - 1;
    }
    for (int i = 0; i < v[2].size(); i++) {
      long long int h = power(n, v[2][i]);
      ans += h - 1;
    }
    for (int i = 0; i < v[3].size(); i++) {
      long long int h = power(n, v[3][i]);
      ans -= h - 1;
    }
    cout << ans << "\n";
  }
}
