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 = (long long)1e9 + 7;
vector<long long> vec, vec2;
bool ck(long long num) {
  long long l = 2, r = min((long long)1e9 + 1, num);
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (mid * mid > num)
      r = mid;
    else
      l = mid;
  }
  return l * l == num;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 2; i <= 1e6; i++) {
    long long tmp = i * 1LL * i;
    while (tmp <= 1e18 / i) {
      tmp *= i;
      vec.push_back(tmp);
    }
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 0; i < (int)vec.size(); i++) {
    if (!ck(vec[i])) vec2.push_back(vec[i]);
  }
  int t;
  long long n;
  cin >> t;
  while (t--) {
    cin >> n;
    long long ans =
        n - 1 - (upper_bound(vec2.begin(), vec2.end(), n) - vec2.begin());
    ans -= (long long)sqrt(n) - 1;
    cout << ans << "\n";
  }
  return 0;
}
