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 i64 = long long;
constexpr i64 OO = i64(1.2e18);
i64 mul(i64 a, i64 b) { return (a <= OO / b ? min(OO, a * b) : OO); }
i64 fpow(i64 a, int b) {
  i64 t = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) t = mul(t, a);
    a = mul(a, a);
  }
  return t;
}
i64 count(i64 n, i64 e) {
  i64 t = pow(n, 1.0 / e);
  while (fpow(t, e) <= n) ++t;
  return t - 2;
}
void solve() {
  i64 n;
  cin >> n;
  vector<i64> cnt(66, 0);
  i64 ans = n - 1;
  for (int e = 60; e >= 2; --e) {
    cnt[e] += count(n, e);
    for (int d = 2; d < e; ++d)
      if (e % d == 0) cnt[d] -= cnt[e];
    ans -= cnt[e];
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
