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 N = 1004006;
const int M = 1000006;
const long long oo = 1.5e18;
long long vet[N];
int sz;
long long sq(long long n) {
  long long ans = max((long long)sqrt((long double)n) - 2, 1LL);
  while ((ans + 1) * (ans + 1) <= n) ans++;
  return ans;
}
long long is_square(long long n) {
  long long x = sq(n);
  return x * x == n;
}
long long solve(long long n) {
  long long ans = n - sq(n);
  ans -= int(upper_bound(vet, vet + sz, n) - vet);
  return ans;
}
int main() {
  for (int i = 2; i < M; i++) {
    long long b = 1LL * i * i * i, lim = oo / i;
    while (1) {
      if (!is_square(b)) vet[sz++] = b;
      if (b > lim) break;
      b *= i;
    }
  }
  sort(vet, vet + sz);
  sz = int(unique(vet, vet + sz) - vet);
  int t;
  long long n;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) {
    scanf("%lld", &n);
    printf("%lld\n", solve(n));
  }
}
