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 long long inf = (long long)1e18 + 1ll;
const int maxn = 100050;
vector<int> G[maxn];
long long f[66];
long double Pow(long long x, long long n) {
  long double ans = 1, base = x;
  while (n) {
    if (n & 1) ans = ans * base;
    base = base * base;
    n >>= 1;
    if (ans >= inf) return inf;
  }
  return ans;
}
long long solve(long long n, int k) {
  long long ans = exp((long double)log((long double)n) / k);
  ans = max(ans - 2, 0ll);
  while (Pow(ans + 1, k) <= n) ans++;
  return ans;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n;
    scanf("%lld", &n);
    for (int i = 60; i >= 1; --i) {
      f[i] = solve(n, i);
      for (int j = i + i; j <= 60; j += i) f[i] -= f[j];
    }
    cout << f[1] + 1 << endl;
  }
  return 0;
}
