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 int power(long long int n, int p) {
  if (p == 1) return n;
  long long int s, t = power(n, p / 2);
  if (t == -1) return -1;
  if (__builtin_mul_overflow(t, t, &s)) return -1;
  if (p % 2 == 0) {
    return s;
  }
  if (__builtin_mul_overflow(s, n, &t)) return -1;
  return t;
}
long long int countFunc(long long int n, int p) {
  long long int tmp = powl(n, 1.0 / p);
  tmp = max(1LL, tmp - 5);
  while (true) {
    long long int t = power(tmp + 1, p);
    if (t == -1 || t > n) break;
    tmp++;
  }
  return tmp;
}
int calcMob(int n) {
  int tmp = 2;
  vector<int> pw;
  while (n > 1) {
    if (n % tmp == 0) {
      int cnt = 0;
      while (n % tmp == 0) {
        n /= tmp;
        cnt++;
      }
      pw.push_back(cnt);
    }
    tmp++;
  }
  for (size_t i = 0; i < pw.size(); i++) {
    if (pw[i] > 1) return 0;
  }
  if (pw.size() % 2 == 0) return 1;
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  int mob[64];
  for (int i = 1; i < 64; i++) {
    mob[i] = calcMob(i);
  }
  while (t--) {
    long long int n;
    cin >> n;
    long long int cnt[64];
    for (int i = 2; i < 64; i++) {
      cnt[i] = countFunc(n, i) - 1;
    }
    cnt[1] = n - 1;
    long long int ans = 0;
    for (int i = 1; i < 64; i++) {
      ans += cnt[i] * mob[i];
    }
    cout << ans << endl;
  }
}
