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;
unsigned long long A[17] = {2,  3,  5,  7,  11, 13, 17, 19, 23,
                            29, 31, 37, 41, 43, 47, 53, 59};
unsigned long long B[17] = {6,  10, 14, 15, 21, 22, 26, 33, 34,
                            35, 38, 39, 46, 51, 55, 57, 58};
unsigned long long power(unsigned long long x, unsigned long long y) {
  unsigned long long a = 1;
  for (unsigned long long j = 0; j < y; j++) a *= x;
  return a;
}
unsigned long long root(unsigned long long A, unsigned long long q) {
  if (q >= 38) {
    if (power(2, q) <= A)
      return 2;
    else
      return 1;
  } else if (q >= 30) {
    if (power(3, q) <= A)
      return 3;
    else if (power(2, q) <= A)
      return 2;
    else
      return 1;
  } else if (q >= 26) {
    if (power(4, q) <= A)
      return 4;
    else if (power(3, q) <= A)
      return 3;
    else if (power(2, q) <= A)
      return 2;
    else
      return 1;
  } else if (q >= 24) {
    if (power(5, q) <= A)
      return 5;
    else if (power(4, q) <= A)
      return 4;
    else if (power(3, q) <= A)
      return 3;
    else if (power(2, q) <= A)
      return 2;
    else
      return 1;
  } else if (q >= 22) {
    if (power(6, q) <= A)
      return 6;
    else if (power(5, q) <= A)
      return 5;
    else if (power(4, q) <= A)
      return 4;
    else if (power(3, q) <= A)
      return 3;
    else if (power(2, q) <= A)
      return 2;
    else
      return 1;
  } else if (q >= 20) {
    if (power(7, q) <= A)
      return 7;
    else if (power(6, q) <= A)
      return 6;
    else if (power(5, q) <= A)
      return 5;
    else if (power(4, q) <= A)
      return 4;
    else if (power(3, q) <= A)
      return 3;
    else if (power(2, q) <= A)
      return 2;
    else
      return 1;
  } else {
    long double s = 1.0, u = s / q;
    unsigned long long i = pow(A, u) - 1;
    while (power(i, q) <= A) i++;
    return i - 1;
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    unsigned long long n, x = 0;
    cin >> n;
    for (int i = 0; i < 17; i++) {
      x += (root(n, A[i]) - 1);
    }
    for (int i = 0; i < 17; i++) {
      x -= (root(n, B[i]) - 1);
    }
    x += (root(n, 30) + root(n, 42) - 2);
    cout << n - 1 - x << "\n";
  }
}
