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;
int T, tot, prime[75], check[75], u[75];
long long n, ans;
const int maxn = 70;
long long Pow(long long a, long long b) {
  long long res = 1;
  long double rr = 1, A = a;
  while (b) {
    if (b & 1) res = res * a, rr = rr * A;
    a = a * a;
    A = A * A;
    b >>= 1;
  }
  if (rr > 1100000000000000000.0) return 1100000000000000000.0;
  return res;
}
long long work(long long x, long long y) {
  long long tmp = (long long)powl(x, 1.0 / y);
  for (;; tmp++)
    if (Pow(tmp, y) > x) return tmp - 1;
}
int main() {
  u[1] = 1;
  for (int i = 2; i <= maxn; i++) {
    if (!check[i]) prime[++tot] = i, u[i] = -1;
    for (int j = 1; j <= tot; j++) {
      if (i * prime[j] > maxn) break;
      check[i * prime[j]] = 1;
      if (i % prime[j])
        u[i * prime[j]] = -u[i];
      else {
        u[i * prime[j]] = 0;
        break;
      }
    }
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%I64d", &n);
    ans = 0;
    for (int i = 1; i <= 60; i++) {
      long long tmp = work(n, i) - 1;
      if (!tmp) break;
      ans += u[i] * tmp;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
