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;
template <typename T>
void read(T &x) {
  char c = getchar();
  x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
}
const int N = 2e6 + 7;
long long a[N], n, t, cnt;
void pre() {
  for (int i = 2; i <= 1e6; i++) {
    long long no = i, pp = i;
    pp *= i;
    no *= pp;
    long long x = trunc(sqrt(i));
    if (x * x == i) continue;
    while (no < 1e18 / pp) a[++cnt] = no, no *= pp;
    a[++cnt] = no;
  }
  sort(a + 1, a + cnt + 1);
  cnt = unique(a + 1, a + cnt + 1) - a - 1;
}
int main() {
  read(t);
  pre();
  while (t--) {
    read(n);
    cout << n - (upper_bound(a + 1, a + cnt + 1, n) - a - 1) -
                (long long)sqrt(n)
         << endl;
  }
  return 0;
}
