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 mul(long long x, long long y) {
  if (x * 1.0 * y > 4e18) return 4e18;
  return x * y;
}
long long pw(long long x, long long y) {
  if (y == 1) return x;
  long long p = pw(x, y >> 1);
  p = mul(p, p);
  if (y & 1) p = mul(p, x);
  return p;
}
long long root(long long x, long long y) {
  long long z = pow(x, 1.0 / y);
  while (pw(z + 1, y) <= x) z++;
  while (pw(z, y) > x) z--;
  return z;
}
int mobius[66];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  fill(mobius + 1, mobius + 66, 1);
  for (int i = 2; i < 66; i++)
    for (int j = 2 * i; j < 66; j += i) mobius[j] -= mobius[i];
  int t;
  cin >> t;
  while (t--) {
    long long x;
    cin >> x;
    long long sol = x;
    for (int i = 2; i <= 66; i++) sol -= mobius[i] * root(x, i);
    cout << sol << '\n';
  }
}
