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 int N = 5e6 + 5;
const long long inf = 1e18;
long long ans[N], tot;
long long great_power(long long p) {
  long long x = (long long)sqrt(p);
  while (x * x < p) ++x;
  while (x * x > p) --x;
  return x;
}
bool issqr(long long p) {
  long long x = great_power(p);
  return x * x != p;
}
void init() {
  for (int i = 2; i <= 1000000; ++i) {
    long long p = 1ll * i * i;
    while (p <= inf / i) {
      p *= i;
      if (issqr(p)) ans[++tot] = p;
    }
  }
}
signed main() {
  init();
  sort(ans + 1, ans + tot + 1);
  tot = unique(ans + 1, ans + tot + 1) - ans - 1;
  int T, res;
  long long n;
  cin >> T;
  while (T--) {
    cin >> n;
    res = upper_bound(ans + 1, ans + tot + 1, n) - ans - 1;
    cout << n - res - great_power(n) << endl;
  }
  return 0;
}
