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 qpow(unsigned long long a, int e) {
  unsigned long long res = 1;
  for (; e; e >>= 1) {
    if (e & 1) res *= a;
    a *= a;
  }
  return res;
}
int main() {
  int T;
  static unsigned long long mx[65];
  mx[2] = 1e9, mx[3] = 1e6, mx[4] = 31622, mx[5] = 3981;
  for (int i = 6; i <= 64; i++) {
    mx[i] = pow(1e18, 1.0 / i);
    while (pow(mx[i], i) < 1e18) mx[i]++;
    while (pow(mx[i], i) > 1e18) mx[i]--;
  }
  cin >> T;
  while (T--) {
    unsigned long long n;
    cin >> n;
    static unsigned long long num[65];
    for (int i = 2; i <= 64; i++) {
      num[i] = pow(n, 1.0 / i);
      num[i] = min(num[i], mx[i]);
      while (num[i] < mx[i] && qpow(num[i], i) < n) num[i]++;
      while (qpow(num[i], i) > n) num[i]--;
      num[i]--;
    }
    num[1] = n - 1;
    for (int i = 64; i >= 1; i--) {
      for (int j = i + i; j <= 64; j += i) {
        num[i] -= num[j];
      }
    }
    cout << num[1] << '\n';
  }
}
