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;
void doRoutine() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
const long long LOGN = 65;
const long long MAXN = 1000000000000000000;
long long max_sqrt[LOGN];
long long fpow(long long a, long long n) {
  if (n == 0) {
    return 1;
  } else if (n % 2 == 0) {
    return fpow(a * a, n / 2);
  } else {
    return fpow(a, n - 1) * a;
  }
}
long long get_root(long long n, long long i) {
  long long l = -1, r = max_sqrt[i] + 1;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    if (fpow(m, i) <= n) {
      l = m;
    } else {
      r = m;
    }
  }
  return l;
}
long long solve(long long n) {
  long long logn = 65;
  vector<long long> dp(logn);
  vector<long long> root(logn);
  for (long long i = 1; i < logn; ++i) {
    root[i] = get_root(n, i);
  }
  for (long long i = logn - 1; i >= 0; --i) {
    if (root[i] >= 2) {
      long long sum = 0;
      for (long long x = 2; x < LOGN; ++x) {
        long long id = i * x;
        if (id < logn) {
          sum += dp[id];
        }
      }
      dp[i] = root[i] - 1 - sum;
    }
  }
  return dp[1];
}
signed main() {
  doRoutine();
  max_sqrt[1] = MAXN;
  max_sqrt[2] = 1000000000;
  for (long long i = 3; i < LOGN; ++i) {
    max_sqrt[i] = (long long)floor(pow((double)MAXN, 1. / (double)i));
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << solve(n) << endl;
  }
  return 0;
}
