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;
long long t;
void read() { cin >> t; }
vector<pair<long long, long long> > q;
vector<long long> nums[64];
long long ans[100003];
long long dp[64];
void solve() {
  for (long long i = 1; i <= t; i++) {
    long long n;
    cin >> n;
    q.push_back({n, i});
  }
  sort(q.begin(), q.end());
  reverse(q.begin(), q.end());
  for (long long i = 3; i <= 60; i++) {
    for (long long j = 2;; j++) {
      long long temp = 1000000000000000000;
      for (long long k = 1; k <= i; k++) temp /= j;
      if (temp == 0) break;
      temp = 1;
      for (long long k = 1; k <= i; k++) temp *= j;
      nums[i].push_back(temp);
    }
  }
  for (auto i : q) {
    dp[2] = sqrt(i.first) - 1;
    for (long long j = 3; j <= 60; j++) {
      while (!nums[j].empty()) {
        if (nums[j].back() <= i.first) break;
        nums[j].pop_back();
      }
      dp[j] = nums[j].size();
    }
    long long res = i.first - 1;
    for (long long j = 60; j >= 2; j--) {
      for (long long k = 2 * j; k <= 60; k += j) dp[j] -= dp[k];
      res -= dp[j];
    }
    ans[i.second] = res;
  }
  for (long long i = 1; i <= t; i++) cout << ans[i] << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  read();
  solve();
}
