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 MaxN = 70 + 7;
const long long inf = 2e18 + 10;
int mu[MaxN];
bool mark[MaxN];
void prep() {
  for (int i = 2; i < MaxN; i++) mu[i] = -1;
  for (int i = 2; i < MaxN; i++) {
    if (mark[i]) continue;
    for (int j = i; j < MaxN; j += i) {
      mark[j] = true;
      if (j % (i * i) == 0)
        mu[j] = 0;
      else
        mu[j] *= -1;
    }
  }
}
long long pw(long long base, long long exp) {
  long long ans = 1;
  for (; exp; exp >>= 1, base = base < inf / base ? base * base : inf)
    if (exp & 1) ans = ans < inf / base ? ans * base : inf;
  return ans;
}
int main() {
  int T;
  scanf("%d", &T);
  prep();
  while (T--) {
    long long n;
    scanf("%lld", &n);
    long long ni = n;
    long long ans = n - 1;
    for (int i = 2; i < MaxN; i++) {
      if (ni == 1) break;
      long long l = 1, r = ni;
      while (l < r) {
        long long mid = l + (r - l + 1) / 2;
        if (pw(mid, i) > n)
          r = mid - 1;
        else
          l = mid;
      }
      ni = l;
      ans -= 1ll * (ni - 1) * mu[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}
