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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
const int ms = 65;
const int mv = 1.1e6;
const long long INF = (long long)1e18;
long long top[mv], cur[mv];
std::vector<long long> down[ms];
int pt[ms];
bool np[ms];
int mob[ms];
void pre() {
  for (int i = 1; i < ms; i++) {
    mob[i] = 1;
  }
  for (int i = 2; i < ms; i++) {
    if (np[i]) continue;
    for (int j = i; j < ms; j += i) {
      np[j] = true;
      mob[j] *= -1;
      if ((j / i) % i == 0) {
        mob[j] = 0;
      }
    }
  }
  for (int i = 1; i < ms; i++) {
  }
  for (int i = 2; i < mv; i++) {
    top[i] = INF;
    top[i] /= i;
    top[i] /= i;
    cur[i] = (long long)i * i;
  }
  for (int j = 3; j < ms; j++) {
    for (int i = 2; top[i] / i > 0; i++) {
      top[i] /= i;
      cur[i] *= i;
      down[j].push_back(cur[i]);
    }
  }
}
long long solve(long long n) {
  long long ans = n - 1;
  for (int i = 3; i < ms; i++) {
    while (pt[i] < (int)down[i].size() && down[i][pt[i]] <= n) {
      pt[i]++;
    }
  }
  long long sqr = (long long)sqrt((double)ans);
  sqr = std::max(sqr - 2, 1LL);
  while ((sqr + 1) * (sqr + 1) <= n) sqr++;
  ans -= sqr - 1;
  for (int i = 3; i < ms; i++) {
    ans += mob[i] * pt[i];
  }
  return ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  pre();
  int t;
  std::cin >> t;
  std::vector<std::pair<long long, int>> qries;
  for (int i = 0; i < t; i++) {
    long long n;
    std::cin >> n;
    qries.emplace_back(n, i);
  }
  std::vector<long long> ans(t, 0);
  std::sort(qries.begin(), qries.end());
  for (auto q : qries) {
    ans[q.second] = solve(q.first);
  }
  for (int i = 0; i < t; i++) {
    std::cout << ans[i] << '\n';
  }
}
