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, mu[61], pri[61], tot;
bool notp[61];
long long n, ans;
inline long double kasumi(long double a, int b) {
  long double res = 1.0;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &t);
  mu[1] = 1;
  notp[0] = notp[1] = true;
  for (register int i = 2; i <= 60; i++) {
    if (!notp[i]) {
      mu[i] = -1;
      pri[tot++] = i;
    }
    for (register int j = 0; j < tot && i * pri[j] <= 60; j++) {
      notp[i * pri[j]] = true;
      if (i % pri[j])
        mu[i * pri[j]] = -mu[i];
      else
        break;
    }
  }
  while (t--) {
    ans = 0;
    scanf("%lld", &n);
    for (register int i = 1; i <= 60; i++) {
      long long tmp = (long long)pow(n, 1.0L / i) + 2;
      while (kasumi(tmp, i) > n) --tmp;
      ans += mu[i] * (tmp - 1);
    }
    printf("%lld\n", ans);
  }
}
