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 inf = 0x3f3f3f3f;
const int N = 105;
long long read() {
  long long s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * f;
}
int cnt, mu[N], vis[N], prime[N];
long double Pow(long double x, long long y) {
  long double ans = 1;
  while (y) {
    if (y & 1) ans = ans * x;
    x = x * x;
    y >>= 1;
  }
  return ans;
}
void Prepare() {
  mu[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      mu[i] = -1;
      prime[++cnt] = i;
    }
    for (int j = 1; j <= cnt && i * prime[j] < N; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j])
        mu[i * prime[j]] = -mu[i];
      else {
        mu[i * prime[j]] = 0;
        break;
      }
    }
  }
}
int main() {
  int Case = read();
  Prepare();
  while (Case--) {
    long long n = read(), ans = 0;
    for (int i = 1; i <= 60; i++) {
      long long tmp = (long long)(pow(n, (long double)(1) / i)) + 1;
      while (Pow(tmp, i) > n) tmp--;
      ans += mu[i] * (tmp - 1);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
