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;
const long long mod = 1e9 + 7;
const double pi = acos(-1);
int n, i, j, k, t;
bool pp[1000002];
long long val[70];
long long limit[] = {
    0,  0,  1000000000, 1000000, 31622, 3981, 1000, 372, 177, 100, 63, 43,
    31, 24, 19,         15,      13,    11,   10,   8,   7,   7,   6,  6,
    5,  5,  4,          4,       4,     4,    3,    3,   3,   3,   3,  3,
    3,  3,  2,          2,       2,     2,    2,    2,   2,   2,   2,  2,
    2,  2,  2,          2,       2,     2,    2,    2,   2,   2,   2,  2,
};
long long powd(long long val, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1) return val;
  long long tmp = powd(val, exp / 2);
  tmp *= tmp;
  if (exp & 1) tmp *= val;
  return tmp;
}
long long root(long long val, long long exp) {
  long long ans = powl(val, 1.0L / exp);
  if (powd(ans, exp) > val) return ans - 1;
  if (ans < limit[exp] && powd(ans + 1, exp) <= val) return ans + 1;
  return ans;
}
long long solve(long long x) {
  memset(val, 0, sizeof val);
  int ma = log2((long double)x);
  long long ans = 0;
  for (int i = ma; i >= 2; --i) {
    val[i] = root(x, i) - 1LL;
    for (int j = 2 * i; j <= ma; j += i) {
      val[i] -= val[j];
    }
    ans += val[i];
  }
  return ans + 1;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    long long val;
    scanf("%lld", &val);
    printf("%lld\n", val - solve(val));
  }
  return 0;
}
