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;
mt19937 rnd(time(0));
const long long INF = 1000000000000000000LL;
long long bin(long long a, long long b) {
  if (!b) return 0;
  if (b % 2) return min(INF + 1, bin(a, b - 1) + a);
  long long t = bin(a, b / 2);
  return min(INF + 1, t + t);
}
int solve() {
  vector<long long> p;
  for (long long i = 1; i * i * i <= INF; i++) {
    long long t = sqrt(i);
    t = max(0LL, t - 3);
    while (t * t < i) t++;
    if (t * t == i) continue;
    long long q = i * i * i;
    long long cur = 3;
    while (q <= INF) {
      if (cur % 2) p.push_back(q);
      q = bin(q, i);
      cur++;
    }
  }
  vector<long long> q;
  sort(p.begin(), p.end());
  q.push_back(p[0]);
  for (int i = 1; i < p.size(); i++)
    if (p[i] != p[i - 1]) q.push_back(p[i]);
  int t;
  cin >> t;
  p.clear();
  for (int i = 0; i < t; i++) {
    long long n;
    cin >> n;
    long long ans = sqrt(n);
    ans = max(0LL, ans - 3);
    while (ans * ans <= n) ans++;
    ans--;
    ans += (long long)(upper_bound(q.begin(), q.end(), n) - q.begin());
    cout << n - ans << '\n';
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  int T = 100;
  for (int i = 1; i <= T; i++) {
    if (solve()) break;
  }
  return 0;
}
