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 q;
long long L, R;
set<long long> s;
long long v2[2000002], x;
void precalculate() {
  long long LLMAX = 1e18;
  for (int i = 2; i <= 1000000; ++i) {
    long long x = 1LL * i * i * i;
    while (1) {
      if ((int)sqrt(x) != sqrt(x)) s.insert(x);
      if (x <= LLMAX / i)
        x = x * i;
      else
        break;
    }
  }
  for (set<long long>::iterator it = s.begin(); it != s.end(); ++it)
    v2[++x] = *it;
  v2[x + 1] = 2e18;
}
long long cb(long long nr) {
  int b = 1;
  int e = x;
  while (b <= e) {
    int m = (b + e) / 2;
    if (v2[m] <= nr && v2[m + 1] > nr) return m;
    if (v2[m] <= nr)
      b = m + 1;
    else
      e = m - 1;
  }
  return 0;
}
long long cb2(long long nr) {
  int b = 0;
  int e = 1e9;
  while (b <= e) {
    int m = (b + e) / 2;
    if (1LL * m * m <= nr && 1LL * (m + 1) * (m + 1) > nr) return m;
    if (1LL * m * m <= nr)
      b = m + 1;
    else
      e = m - 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> q;
  precalculate();
  for (int i = 1; i <= q; ++i) {
    cin >> R;
    L = 1;
    long long sol = cb(R) - cb(L - 1) + cb2(R) - cb2(L - 1);
    sol = R - sol;
    cout << sol << '\n';
  }
  return 0;
}
