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;
using ll = long long;
using ld = long double;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ll T;
  cin >> T;
  vector<ll> p(64);
  for (ll d = 1; d < 63; d++) {
    ll x = d;
    for (ll i = 2; i < d + 1; i++) {
      if (!(x % i)) {
        x /= i;
        if (!(x % i)) {
          p[d] = -1;
          break;
        }
        p[d]++;
      }
    }
  }
  for (ll z = 0; z < T; z++) {
    ld n;
    cin >> n;
    ll ans = 0;
    for (ll d = 1; d < 63; d++) {
      for (ll j = floor(pow(n, 1.0L / d)); j < 2e18; j++) {
        if (pow(ld(j), ld(d)) > ld(n)) {
          if (p[d] != -1) {
            if (p[d] % 2) {
              ans -= j - 2;
            } else {
              ans += j - 2;
            }
          }
          break;
        }
      }
    }
    cout << ans << endl;
  }
}
