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 P = 64;
long long n;
int mu[P + 5], vis[P + 5], p[P + 5], tot = 0;
void sieve(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      p[++tot] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= tot && p[j] * i <= n; j++) {
      vis[i * p[j]] = 1;
      if (i % p[j] == 0) break;
      mu[i * p[j]] = -mu[i];
    }
  }
  return;
}
long long calc(long long x, int y) {
  long long res = 1;
  for (; y; y >>= 1, x = x * x)
    if (y & 1) res = res * x;
  return res;
}
long long check(long long x, int y, long long k) {
  long long res = 1;
  for (int i = 1; i <= y; i++) {
    if (res > k / x) return 1;
    res *= x;
  }
  return (res > k);
}
long long Sq(long long x, int y) {
  long long k = pow((long double)x, (long double)1.0 / y);
  while (!check(k + 1, y, n)) k++;
  return k;
}
void work() {
  scanf("%lld", &n);
  long long ans = 0;
  for (int i = 1; i <= P; i++) {
    ans += mu[i] * (Sq(n, i) - 1);
  }
  printf("%lld\n", ans);
  return;
}
int main() {
  sieve(P);
  int t;
  scanf("%d", &t);
  for (; t--;) work();
  return 0;
}
