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 int N = 1e6;
const long long int inf = 1e18;
vector<long long int> v, arr;
void pf() {}
long long int sqrtfun(long long int num) {
  long long int lo = 0, hi = (long long int)2e9, mid;
  if (num == 1) return num;
  while (lo + 1 < hi) {
    mid = lo + hi >> 1LL;
    if (mid * mid == num)
      return mid;
    else if (mid * mid > num)
      hi = mid;
    else
      lo = mid;
  }
  return (mid * mid > num) ? mid - 1 : mid;
}
long long int get(long long int r) {
  int ans = upper_bound(arr.begin(), arr.end(), r) - arr.begin();
  return ans + sqrtl(r) - 1;
}
void solve() {
  for (long long int i = 2; i <= N; i++) {
    for (long long int j = i * i * i; j <= inf; j *= i) {
      long long int x = sqrtl(j);
      if (x * x != j) v.push_back(j);
      if (inf / i < j) break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, l, r, ans = 0;
  cin >> t;
  solve();
  sort(v.begin(), v.end());
  for (auto i : v) {
    if (arr.empty())
      arr.push_back(i);
    else if (arr.back() != i)
      arr.push_back(i);
  }
  while (t--) {
    cin >> r;
    cout << r - get(r) - 1 << '\n';
  }
  return 0;
}
