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;
vector<ll> noneleg;
ll pw(ll a, ll e) {
  ll ans = 1;
  while (e) {
    if (e & 1) ans *= a;
    e /= 2;
    a = a * a;
  }
  return ans;
}
bool isx2(ll x) {
  ll tmp = round(sqrt(x));
  return tmp * tmp == x;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 3; i <= 64; ++i) {
    ll mx = (ll)floor(pow(1E18, 1.0 / i) + 0.01);
    for (ll x = 2; x <= mx; ++x) {
      ll tmp = pw(x, i);
      if (!isx2(tmp)) {
        noneleg.push_back(tmp);
      }
    }
  }
  sort(noneleg.begin(), noneleg.end());
  noneleg.erase(unique(noneleg.begin(), noneleg.end()), noneleg.end());
  ll T, N;
  cin >> T;
  while (T--) {
    cin >> N;
    ll x2 = floor(sqrt(N));
    if (N == (x2 + 1) * (x2 + 1)) x2++;
    ll others =
        upper_bound(noneleg.begin(), noneleg.end(), N) - noneleg.begin();
    cout << N - x2 - others << '\n';
  }
}
