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 int mu[61];
void pre() {
  vector<bool> isPrime(1000, true);
  for (long long int i = 2; i < 60; i++) {
    if (isPrime[i]) {
      for (long long int j = i * i; j < 1000; j += i) isPrime[j] = false;
    }
  }
  mu[1] = 1;
  for (long long int i = 2; i < 60; i++) {
    int cnt = 0;
    for (long long int j = 2; j < 60; j++) {
      if (isPrime[j] && i % j == 0) {
        if (i % (j * j) == 0) {
          cnt = -1;
          break;
        } else
          cnt++;
      }
    }
    if (cnt == -1) {
      mu[i] = 0;
    } else {
      mu[i] = (cnt % 2 == 0 ? 1 : -1);
    }
  }
}
long long int expo(long long int x, long long int y) {
  long long int ans = 1;
  while (y > 0) {
    if ((long double)ans * (long double)x > 1e18) {
      return (long long int)1e18 + 1ll;
    }
    if (y & 1) ans *= x;
    if ((long double)x * (long double)x > 1e18 && y > 1) {
      return (long long int)1e18 + 1ll;
    }
    x *= x;
    y >>= 1;
  }
  return ans;
}
void solve() {
  long long int n;
  cin >> n;
  long long int ans = 0;
  for (long long int i = 1; i < 61; i += 1) {
    long long int l = 1, r = expo(10ll, (18 + i - 1) / i) + 1ll;
    while (l + 1 < r) {
      long long int mid = (l + r) / 2;
      if (expo(mid, i) > n) {
        r = mid;
      } else {
        l = mid;
      }
    }
    ans += (l - 1) * mu[(long long int)i];
  }
  cout << ans << '\n';
}
int main() {
  expo(10, 18);
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(10);
  cout << fixed;
  pre();
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
