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 = 1e6 + 7;
bool Prime[N];
int p[N >> 3], mu[N], tot;
void minit(int x) {
  mu[1] = 1;
  for (int i = 2, iE = x; i <= iE; i++) {
    if (!Prime[i]) p[++tot] = i, mu[i] = -1;
    for (int j = 1; p[j] * i <= x && j <= tot; j++) {
      Prime[p[j] * i] = 1;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      }
      mu[i * p[j]] = -mu[i];
    }
  }
}
long long n, f[N], ans;
long double qpow(long double x, int y) {
  long double res = 1;
  for (; y; x = x * x, y >>= 1)
    if (y & 1) res *= x;
  return res;
}
long long get(int x) {
  long long l = floor(pow(n, 1.0L / x)) - 1, mid = l + 1, r = l + 2;
  if (qpow(r, x) <= (long double)n) return r;
  if (qpow(mid, x) <= (long double)n) return mid;
  return l;
}
void Main() {
  ans = 0;
  scanf("%lld", &n);
  for (int i = 59, iE = 1; i >= iE; i--) f[i] = get(i) - 1, ans += f[i] * mu[i];
  printf("%lld\n", ans);
}
int main() {
  minit(1111);
  int T;
  scanf("%d", &T);
  while (T--) Main();
  return 0;
}
