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 int maxn = 1e6 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const int mod = 1e9 + 7;
vector<long long> v;
long long first(long long x) {
  return (x ? upper_bound(v.begin(), v.end(), x) - v.begin() + sqrt(x) : 0);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 2; i <= 1e6; i++) {
    long long ret = i;
    while (ret <= 1e18 && log10(ret) + log10(i) <= 18) {
      ret *= i;
      long long s = sqrt(ret);
      if (ret != s * s) v.push_back(ret);
    }
  }
  sort(v.begin(), v.end());
  v.erase(unique((v).begin(), (v).end()), v.end());
  long long q;
  cin >> q;
  while (q--) {
    long long x;
    cin >> x;
    cout << (x - 1) - (first(x) - first(1)) << endl;
  }
}
