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 = 65;
bool vis[maxn];
int mu[maxn];
int prim[maxn];
int cnt;
void init() {
  memset(vis, 0, sizeof(vis));
  memset(mu, 0, sizeof(mu));
  mu[1] = 1;
  for (int i = 2; i <= maxn - 1; ++i) {
    if (!vis[i]) {
      prim[++cnt] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= cnt && 1ll * prim[j] * i < maxn; ++j) {
      vis[prim[j] * i] = 1;
      if (i % prim[j] == 0)
        break;
      else
        mu[prim[j] * i] = -mu[i];
    }
  }
}
int main() {
  init();
  int caseCnt;
  scanf("%d", &caseCnt);
  while (caseCnt--) {
    long long n;
    scanf("%lld", &n);
    long long ans = n - 1;
    for (int i = 2; i < maxn; ++i) {
      long long x =
          (long long)powl((long double)n + 0.3, (long double)1.0 / i) - 1;
      ans += x * mu[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}
