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 INF = 1e9;
const int maxn = 1e6 + 50;
const long long mod = 1e9 + 7;
int prime[150];
bool notprime[150];
int mu[150];
void getMobius(int N) {
  mu[1] = 1;
  int tot = 0;
  for (int i = 2; i <= N; i++) {
    if (!notprime[i]) {
      prime[++tot] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= tot; j++) {
      if (i * prime[j] > N) break;
      notprime[i * prime[j]] = true;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      } else {
        mu[i * prime[j]] = -mu[i];
      }
    }
  }
}
long double ksm(long double a, long long b) {
  long double res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long get(long long x, long long i) {
  long long tmp = powl(x, 1.0 / i) + 2;
  while (ksm(tmp, i) > x) --tmp;
  return tmp;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  getMobius(120);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long ans = n - 1;
    for (int i = 2; i <= 60; i++) {
      ans += mu[i] * (get(n, i) - 1);
    }
    cout << ans << endl;
  }
  return 0;
}
