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 long long INF = (1e18);
bool isSquare(long long x) {
  long long y = (long long)sqrt(x);
  if (y * y == x) return true;
  if ((y - 1) * (y - 1) == x) return true;
  if ((y + 1) * (y + 1) == x) return true;
  return false;
}
bool prime(long long x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
long long f(long long x, long long y) {
  long long ans = 1;
  for (int i = 0; i < y; i++) {
    if (ans > INF / x) return INF + 1;
    ans *= x;
  }
  return ans;
}
set<long long> used;
vector<long long> G;
void init() {
  for (long long i = 3; i <= 64; i += 2) {
    for (long long j = 2;; j++) {
      if (isSquare(j)) continue;
      long long val = f(j, i);
      if (val > INF) break;
      if (used.find(val) != used.end()) continue;
      G.push_back(val);
      used.insert(val);
    }
  }
  sort(G.begin(), G.end());
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  init();
  int t;
  cin >> t;
  while (t--) {
    long long x;
    cin >> x;
    long long ans = (long long)sqrt(x);
    if (ans * ans > x) {
      ans--;
    }
    long long p = lower_bound(G.begin(), G.end(), x + 1) - G.begin();
    ans += p;
    cout << x - ans << '\n';
  }
  return 0;
}
