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 get_sqrt(long long n) {
  long long ini = 1, fim = 1000000000, ans;
  while (ini <= fim) {
    long long mid = (ini + fim) / 2ll;
    if (mid * mid > n)
      fim = mid - 1;
    else
      ans = mid, ini = mid + 1;
  }
  return ans;
}
long long pot[1000010];
bool sq[1000010];
int main(void) {
  vector<long long> V;
  for (int i = 1; i <= 1000000; i++) {
    pot[i] = 1ll * i * i;
    if (pot[i] <= 1000000ll) sq[pot[i]] = 1;
  }
  for (int k = 3; k < 70; k++) {
    for (int i = 2; i <= 1000000 && 1ll * pot[i] * i <= 1000000000000000000ll;
         i++) {
      pot[i] = 1ll * i * pot[i];
      if (k & 1 && !sq[i]) V.push_back(pot[i]);
    }
  }
  sort(V.begin(), V.end());
  V.erase(unique(V.begin(), V.end()), V.end());
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    long long n;
    scanf("%lld", &n);
    long long ans = n - get_sqrt(n);
    ans -= (long long)((upper_bound(V.begin(), V.end(), n) - V.begin()));
    printf("%lld\n", ans);
  }
}
