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;
long long inf = 1e9 + 7;
long long k[60];
void precalc() {
  int i, j;
  for (i = 2; i < 60; i++) {
    k[i]++;
    for (j = 2; j < i; j++)
      if (i % j == 0) k[i] -= k[j];
  }
}
long long Pow(long long x, long long p) {
  long long ans = 1;
  for (long long i = 1; i <= p; i <<= 1) {
    if (i & p) ans *= x;
    x *= x;
  }
  return ans;
}
long long get(long long x) {
  int i;
  long long sqr, ans = 0;
  for (i = 2; i < 60; i++) {
    sqr = (long long)(pow(x, 1.0 / (long double)i) + 1e-4);
    while (Pow(sqr, i) > x) sqr--;
    if (sqr == 1) break;
    ans += (sqr - 1) * k[i];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  precalc();
  int t;
  long long n;
  cin >> t;
  while (t--) {
    cin >> n;
    cout << n - 1 - get(n) << '\n';
  }
}
