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;
int T;
signed long long N;
vector<signed long long> C;
signed long long issq(signed long long V) {
  signed long long a = sqrt(V);
  if (a * a == V) return 1;
  if ((a - 1) * (a - 1) == V) return 1;
  if ((a + 1) * (a + 1) == V) return 1;
  return 0;
}
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  for (signed long long a = 2; a <= 1000000; a++) {
    signed long long v = a * a * a;
    while (v <= 1000000000000000000LL) {
      if (!issq(v)) C.push_back(v);
      if ((double)v * a > 2e18) break;
      v *= a;
    }
  }
  sort((C.begin()), (C.end()));
  C.erase(unique((C.begin()), (C.end())), C.end());
  cin >> T;
  while (T--) {
    cin >> N;
    signed long long ret =
        N - (signed long long)sqrt(N + 0.5) -
        (lower_bound((C.begin()), (C.end()), N + 1) - C.begin());
    cout << ret << endl;
  }
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}
