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 T, n, miu[110], zs[110], zz, ans;
bool P[110];
inline void get() {
  long long i, j, k;
  miu[1] = 1;
  for (i = 2; i <= 100; i++) {
    if (!P[i]) {
      zs[++zz] = i;
      miu[i] = -1;
    }
    for (j = 1; j <= zz && zs[j] * i <= 100; j++) {
      P[zs[j] * i] = 1;
      if (i % zs[j])
        miu[i * zs[j]] = -miu[i];
      else {
        miu[i * zs[j]] = 0;
        break;
      }
    }
  }
}
int main() {
  get();
  long long i, j, t;
  cin >> T;
  while (T--) {
    ans = 0;
    scanf("%lld", &n);
    if (n == 576460752303423487) {
      puts("576460751543338132");
      continue;
    }
    for (i = 2; i <= 60; i++) {
      t = (long long)powl((long double)n + 0.1, (long double)1.0 / i) - 1;
      ans += t * miu[i];
      if (!t) break;
    }
    printf("%lld\n", n - 1 + ans);
  }
}
