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 = 1000005;
const long long INF = 1e18;
int T;
long long val[N * 10];
int tot;
long long mul(long long a, long long b) {
  __int128 c = (__int128)a * b;
  if (c > INF)
    return INF + 1;
  else
    return c;
}
void init(int n = 1000000) {
  for (int i = 2; i <= n; i++) {
    long long p = 1LL * i * i * i;
    while (p <= INF) {
      auto check = [=](long long x) {
        long long t = (long long)sqrt(x);
        while (t * t < x) t++;
        while (t * t > x) t--;
        return t * t != x;
      };
      if (check(p)) val[++tot] = p;
      p = mul(p, i);
    }
  }
  sort(val + 1, val + tot + 1);
  tot = unique(val + 1, val + tot + 1) - val - 1;
  return;
}
long long n;
void solve() {
  scanf("%lld", &n);
  long long t = sqrt(n);
  while (t * t < n) t++;
  while (t * t > n) t--;
  long long ans = (n - (upper_bound(val + 1, val + tot + 1, n) - val - 1) - t);
  printf("%lld\n", ans);
  return;
}
int main() {
  init();
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
