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 N = 1e6 + 5;
const long long maxi = 1e18 + 7;
vector<long long> pw[66];
int64_t f[66];
void prepare() {
  for (long long i = 2, _c = N; i <= _c; ++i) {
    long long now = 1ll * i * i, cnt = 2;
    for (long long j = 3;; j++) {
      if (maxi / now < i) break;
      now *= i;
      if (j > 2) pw[j].push_back(now);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  prepare();
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    long long ans = n - 1;
    for (long long i = 63 - __builtin_clzll(n), _c = 2; i >= _c; --i) {
      if (i == 2)
        f[i] = sqrt(n) - 1;
      else
        f[i] = distance(begin(pw[i]), upper_bound(begin(pw[i]), end(pw[i]), n));
      for (long long j = 2 * i; j <= 63 - __builtin_clzll(n); j += i)
        f[i] -= f[j];
      ans -= f[i];
    }
    cout << ans << '\n';
  }
}
