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 mob[61] = {0, 1,  -1, -1, 0, -1, 1, -1, 0, 0,  1,  -1, 0, -1, 1,  1,
               0, -1, 0,  -1, 0, 1,  1, -1, 0, 0,  1,  0,  0, -1, -1, -1,
               0, 1,  1,  1,  0, -1, 1, 1,  0, -1, -1, -1, 0, 0,  1,  -1,
               0, 0,  0,  1,  0, -1, 0, 1,  0, 1,  1,  -1, 0};
int main() {
  int N;
  scanf("%d\n", &N);
  for (int i = 0; i < N; ++i) {
    long long x;
    scanf("%I64d\n", &x);
    long long res = x - 1;
    for (double j = 2; j <= 60; ++j) {
      long long cur = (long long)floor(pow(x, (1 / j))) + 2;
      while (cur--) {
        long long temp = 1;
        bool tb = true;
        for (int k = 0; k < j; ++k) {
          if (temp > x / cur) {
            tb = false;
            break;
          }
          temp *= cur;
        }
        if (tb) break;
      }
      --cur;
      res += cur * mob[(int)j];
    }
    printf("%I64d\n", res);
  }
}
