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;
inline long long read() {
  long long x = 0;
  char c = getchar();
  bool f = 0;
  for (; c > '9' || c < '0'; f = c == '-', c = getchar())
    ;
  for (; c >= '0' && c <= '9'; x = (x << 1) + (x << 3) + c - '0', c = getchar())
    ;
  return f ? -x : x;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + 48);
}
long long num[3000010], lim = 1000000000000000000ll;
int main() {
  for (int i = 2; i <= 1000000; ++i)
    for (long long x = (long long)i * i * i;; x *= i) {
      long long y = sqrt(x);
      if (y * y != x) num[++*num] = x;
      if (x > lim / i) break;
    }
  sort(num + 1, num + *num + 1);
  *num = unique(num + 1, num + *num + 1) - num - 1;
  int Test = read();
  for (int tt = 1; tt <= Test; ++tt) {
    long long n = read();
    write(n - (upper_bound(num + 1, num + *num + 1, n) - num - 1) -
          (long long)sqrt(n));
    puts("");
  }
}
