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 max1 = 1000000000000000007;
long long power(long long a, long long x) {
  long long ans = 1;
  while (x > 0) {
    if (((1e18 + 3) / (long double)(a)) < ans) return (max1);
    if (x & 1) {
      ans *= a;
    }
    x >>= 1;
    if (x <= 0) break;
    if (((1e18 + 3) / (long double)(a)) < a) return (max1);
    a *= a;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(25);
  long long t, n, a, b, c, d, i, j, k;
  long long cou[65];
  cin >> t;
  while (t--) {
    cin >> n;
    for (i = 0; i < 65; i++) cou[i] = 0;
    for (i = 64; i >= 1; i--) {
      if (i != 1)
        a = pow(n, (long double)((long double)1.0 / (i * (long double)1.0)));
      else
        a = n;
      c = 1;
      for (k = max(a + 1, 1ll); k > 1; k--) {
        b = power(k, i);
        if (b <= n) {
          c = k;
          break;
        }
      }
      a = c;
      cou[i] = a - 1;
      for (j = 2 * i; j < 65; j += i) {
        cou[i] -= cou[j];
      };
    }
    cout << cou[1] << "\n";
  }
  return 0;
}
