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;
long long SQR(long long x) {
  long long mn = 1, mx = min((1LL << 30), x);
  long long ans = 1;
  while (mn <= mx) {
    long long md = (mn + mx) / 2;
    if (md * md > x)
      mx = md - 1;
    else
      ans = md, mn = md + 1;
  }
  return ans;
}
long long MAX = 1e18;
int main() {
  vector<long long> a;
  for (int i = 2; i <= 1000000; i++) {
    long long p = 1LL * i * i;
    while (p <= MAX / i) {
      p *= i;
      long long s = SQR(p);
      if (s * s != p) a.push_back(p);
    }
  }
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
  int T;
  scanf("%d", &T);
  while (T--) {
    long long n;
    scanf("%I64d", &n);
    long long ans = upper_bound(a.begin(), a.end(), n) - a.begin();
    ans += SQR(n);
    printf("%I64d\n", n - ans);
  }
}
