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 <class t>
inline t read(t &x) {
  char c = getchar();
  bool f = 0;
  x = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
template <class t, class... A>
inline void read(t &x, A &...a) {
  read(x);
  read(a...);
}
template <class t>
inline void write(t x) {
  if (x < 0)
    putchar('-'), write(-x);
  else {
    if (x > 9) write(x / 10);
    putchar('0' + x % 10);
  }
}
const long long N = 1e7 + 5;
long long a[N], n;
long long ssss(long long x) {
  long long a = sqrt(x);
  while (a * a < x) a++;
  while (a * a > x) a--;
  return a;
}
void doit() {
  long long x;
  read(x);
  write(x - (upper_bound(a + 1, a + 1 + n, x) - a - 1) - (long long)ssss(x));
  puts("");
}
bool notsqr(long long x) {
  long long a = ssss((long long)x);
  return a * a < x;
}
signed main() {
  for (long long i = 2; i <= 1000000; i++) {
    long long p = i * i;
    while (p <= (1000000000000000000ll) / i) {
      p *= i;
      if (notsqr(p)) a[++n] = p;
    }
  }
  sort(a + 1, a + 1 + n);
  n = unique(a + 1, a + 1 + n) - a - 1;
  long long t;
  read(t);
  while (t--) doit();
}
