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;
int mu[100];
int pot[100];
int flag;
long long maxn;
long long pow(long long b, int e) {
  long long r = 1;
  flag = 0;
  for (int i = 0; i < e; i++) {
    if (r > maxn / b) {
      flag = 1;
      break;
    }
    r = r * b;
  }
  return r;
}
int main() {
  mu[1] = 1;
  for (int x = 1; x < 100; x++) {
    for (int y = 2 * x; y < 100; y += x) {
      mu[y] -= mu[x];
    }
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &maxn);
    long long ans = maxn - 1;
    for (int x = 2; x < 60; x++) {
      long long qt = pow(maxn, 1.0 / x);
      while (pow(qt + 1, x) <= maxn && !flag) qt++;
      ans += mu[x] * (qt - 1);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
