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 re = 0, flag = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    re = (re << 1) + (re << 3) + ch - '0', ch = getchar();
  return re * flag;
}
set<long long> s, ss;
long long x[2000010], y[2000010], cnt1 = 0, cnt2 = 0;
int main() {
  long long T = read(), n, i, j, lim, times = 0;
  for (j = 3; j <= 64; j++) {
    lim = ((long long)pow((long double)1e18, 1.0 / (long double)(j)));
    if (j == 9) lim++;
    for (i = 2; i <= lim; i++) {
      s.insert((long long)((long double)pow((long double)i, (long double)j)));
      if (j % 2 == 0)
        ss.insert(
            (long long)((long double)pow((long double)i, (long double)j)));
    }
  }
  set<long long>::iterator it;
  for (it = s.begin(); it != s.end(); it++) x[++cnt1] = (*it);
  for (it = ss.begin(); it != ss.end(); it++) y[++cnt2] = (*it);
  while (T--) {
    n = read();
    times++;
    long long len3 = upper_bound(x + 1, x + cnt1 + 1, n) - x - 1;
    long long tmp = sqrt((long double)n);
    long long len2 = upper_bound(y + 1, y + cnt2 + 1, n) - y - 1;
    len2 = tmp - 1 - len2;
    cout << n - len3 - len2 - 1 << '\n';
  }
}
