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 = 100005;
long long n;
int cnt[N];
bool ignored[N];
bool lower(long long w, int p) {
  if (w > n) {
    return false;
  }
  long long val = n;
  long long val2 = w;
  for (int i = 2; i <= p; i++) {
    if (val2 <= 1e9) {
      val2 *= w;
    } else {
      val /= w;
    }
    if (val < val2) {
      return false;
    }
  }
  return true;
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 2; i <= 64; i++) {
    if (cnt[i] == 0) {
      for (int j = i; j <= 64; j += i) {
        cnt[j]++;
      }
    }
    int w = i * i;
    while (w <= 64) {
      ignored[w] = true;
      for (int j = 2 * w; j <= 64; j += w) {
        ignored[j] = true;
      }
      w *= i;
    }
  }
  while (t--) {
    scanf("%lld", &n);
    long long ans = 0;
    int last = 1e9;
    for (int j = 2; j <= 64; j++) {
      if (ignored[j]) continue;
      int low = 2;
      int high = last;
      int res = pow(n, 1.0 / j);
      while (pow(res + 1, j) <= n) res++;
      while (pow(res, j) > n) res--;
      last = res;
      if (cnt[j] % 2 == 0) {
        ans -= res - 1;
      } else {
        ans += res - 1;
      }
    }
    printf("%lld\n", n - ans - 1);
  }
  return 0;
}
