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;
vector<int> v;
const double eps = 0.000001;
long long i0, n;
bool ispr(int num) {
  if (num == 1) return 0;
  if (num == 2 || num == 3) return 1;
  if (num % 6 != 1 && num % 6 != 5) return 0;
  int tmp = sqrt(num);
  for (int i = 5; i <= tmp; i += 6)
    if (num % i == 0 || num % (i + 2) == 0) return 0;
  return 1;
}
long long dfs(int p, int num, int k, int r) {
  long long cnt = 0, tmp;
  if (k == 0) {
    tmp = (long long)(pow(n, 1.0 / num) + eps);
    if ((long long)(pow(tmp, num) + eps) > n)
      tmp -= 2;
    else
      tmp--;
    cnt += tmp > 0 ? tmp * ((r % 2) ? 1 : -1) : 0;
  }
  if (k && p < 17) {
    if (num * v[p] < 60) cnt += dfs(p + 1, num * v[p], k - 1, r);
    cnt += dfs(p + 1, num, k, r);
  }
  return cnt;
}
int main() {
  for (int i = 2; i <= 100; i++)
    if (ispr(i)) v.push_back(i);
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    printf("%lld\n",
           n - 1 - dfs(0, 1, 1, 1) - dfs(0, 1, 2, 2) - dfs(0, 1, 3, 3));
  }
  return 0;
}
