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;
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  set<ll> st;
  vector<int> issq(1000001);
  for (int i = 1; i <= 1000; i++) issq[i * i] = 1;
  for (ll i = 2; i <= 1000000; i++) {
    if (issq[i]) continue;
    ll cur = i;
    ll sqi = i * i;
    while (1) {
      if ((double)cur * sqi > 2e18) break;
      cur *= sqi;
      st.insert(cur);
    }
  }
  vector<ll> ngs(st.size());
  int i = 0;
  for (auto e : st) ngs[i++] = e;
  int t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    ll ng3 = upper_bound(ngs.begin(), ngs.end(), n) - ngs.begin();
    ll ng2 = (int)sqrt(n) + 2;
    while (ng2 * ng2 > n) --ng2;
    cout << n - ng3 - ng2 << "\n";
  }
  return 0;
}
