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 inf = 4e18;
long long mul(long long x, long long y) {
  if (x > inf / y) return inf;
  return x * y;
}
long long pw(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = mul(ret, x);
    x = mul(x, x);
    y >>= 1;
  }
  return ret;
}
long long f[61];
signed main() {
  ios_base::sync_with_stdio(false);
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    for (long long i = 60; i >= 1; --i) {
      f[i] =
          (long long)((long double)pow((long double)n, (long double)(1.0 / i)) +
                      (long double)1.0);
      while (pw(f[i], i) > n) f[i]--;
      f[i]--;
      for (long long j = i + i; j <= 60; j += i) f[i] -= f[j];
    }
    cout << f[1] << '\n';
  }
}
