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>
const int maxn = 70;
using namespace std;
vector<int> prime;
int mu[maxn];
bool vis[maxn];
void make_table() {
  mu[1] = 1;
  for (int i = 2; i < maxn; i++) {
    if (!vis[i]) {
      mu[i] = -1;
      prime.push_back(i);
    }
    for (int j = 0; j < (int)prime.size() && i * prime[j] < maxn; j++) {
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else
        mu[i * prime[j]] = -mu[i];
    }
  }
}
long long n;
long long solve() {
  long long ans = 0;
  for (int i = 1; i <= 60; i++) {
    if (!mu[i]) continue;
    ans += 1ll * mu[i] *
           ((long long)powl((long double)n + 0.1, (long double)1.0 / i) - 1);
  }
  return ans;
}
int main() {
  make_table();
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%I64d", &n);
    printf("%I64d\n", solve());
  }
  return 0;
}
