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;
int prs[600], miu[600];
long long fec[62];
bool is_p[600];
void niu_table(long long n) {
  long long p = 0;
  memset(is_p, true, sizeof(is_p));
  memset(miu, 0, sizeof(miu));
  is_p[1] = is_p[0] = false;
  miu[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (is_p[i]) {
      prs[p++] = i;
      miu[i] = -1;
    }
    for (int j = 0; j < p; j++) {
      if (prs[j] * i > n) break;
      is_p[prs[j] * i] = false;
      miu[i * prs[j]] = i % prs[j] ? -miu[i] : 0;
      if (i % prs[j] == 0) break;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<long long> ans;
  niu_table(500);
  int T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    fec[1] = n;
    for (long long i = 2; i <= 60; i++) {
      long double res = powl(n + 0.1, 1.0 / i);
      fec[i] = (long long)res;
    }
    long long r = 0;
    for (long long i = 1; i <= 60; i++) {
      r += miu[i] * (fec[i] - 1);
    }
    ans.emplace_back(r);
    memset(fec, 0, sizeof(fec));
  }
  for (auto &ele : ans) {
    cout << ele << '\n';
  }
  return 0;
}
