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;
long long c[10000010], cc, cd;
const long long mod = 1e18 + 5;
int main() {
  for (int i = 3; i < 60; ++i) {
    bool en = false;
    for (int j = 2;; ++j) {
      long long now = j;
      for (int k = 1; k < i; ++k) {
        if (mod / now < j) {
          en = true;
          break;
        }
        now *= j;
      }
      if (en) break;
      c[cc++] = now;
    }
  }
  sort(c, c + cc);
  cc = unique(c, c + cc) - c;
  for (int i = 0; i < cc; ++i) {
    bool ch = true;
    long long ka = sqrt(c[i]);
    if (ka * ka == c[i] || (ka - 1) * (ka - 1) == c[i] ||
        (ka + 1) * (ka + 1) == c[i])
      ch = false;
    if (ch) c[cd++] = c[i];
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n;
    scanf("%I64d", &n);
    long long s = 1, e = 1e9 + 7, mid;
    while (s < e) {
      mid = e - (e - s) / 2;
      if (mid * mid <= n) {
        s = mid;
      } else {
        e = mid - 1;
      }
    }
    int pos = upper_bound(c, c + cd, n) - c;
    printf("%I64d\n", n - pos - s);
  }
  return 0;
}
