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;
const int N = 100;
const int prime[15] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47};
const double eps = 1e-3;
long long f(long long n, long long d) {
  long long t = powl(n, 1. / d) - eps;
  return t + (powl(t + 1, d) - eps <= n);
}
int mu[100];
int Calcmu(int n) {
  int sgn = 1;
  for (int i = 0, p; i < 4; ++i)
    if (n % (p = prime[i]) == 0) {
      n /= p;
      sgn *= -1;
      if (n % p == 0) return 0;
    }
  if (n > 1) sgn *= -1;
  return sgn;
}
int main() {
  for (int i = 1; i <= 60; ++i) mu[i] = Calcmu(i);
  int Tes;
  scanf("%d", &Tes);
  while (Tes--) {
    long long n;
    scanf("%lld", &n);
    long long ans = 0;
    for (int i = 1; i < 60; ++i) {
      ans += (f(n, i) - 1) * mu[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}
