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;
int q;
long long L, R;
vector<long long> vec, vec2;
long long solve(long long x);
set<int> st;
void precalc() {
  for (long long i = 2; i <= 1e6; i++) {
    for (long long j = i * i * i; j <= 1e18; j *= i) {
      vec.push_back(j);
      if (j > (1e18) / i) break;
    }
  }
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  for (long long i = 0; i < (long long)vec.size(); i++) {
    long long tmp = solve(vec[i]);
    if (tmp * tmp == vec[i]) continue;
    vec2.push_back(vec[i]);
  }
  sort(vec2.begin(), vec2.end());
}
long long solve(long long x) {
  long long l = 0, r = 1e9 + 1;
  while (r - l > 1) {
    long long k = (l + r) / 2;
    if (k * k > x)
      r = k;
    else
      l = k;
  }
  return l;
}
int main() {
  cin >> q;
  precalc();
  while (q--) {
    scanf("%I64d", &R);
    printf("%I64d\n", R - (upper_bound(vec2.begin(), vec2.end(), R) -
                           vec2.begin() + solve(R)));
  }
  return 0;
}
