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 stress();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  stress();
}
long long mx = 1e18;
long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long solve(long long n) {
  map<long long, long long> m;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      while (n % i == 0) {
        m[i]++;
        n /= i;
      }
    }
  }
  if (n > 1) m[n]++;
  long long g = 0;
  for (auto it : m) {
    g = gcd(g, it.second);
  }
  return g;
}
long long naive(long long n) {
  long long sum = 0;
  for (long long i = 2; i <= n; i++) {
    if (solve(i) == 1) sum++;
  }
  return sum;
}
vector<long long> a;
long long best(long long n) {
  long long good = n - 1;
  long long sq = sqrt(n);
  while (sq * sq <= n) sq++;
  while (sq * sq > n) {
    sq--;
  }
  good -= sq - 1;
  auto it = upper_bound(a.begin(), a.end(), n);
  if (it != a.begin()) {
    --it;
    good -= it - a.begin() + 1;
  }
  return good;
}
void stress() {
  set<long long> all;
  for (long long i = 2; i <= 1e6; i++) {
    long long x = i * i * i;
    while (true) {
      all.insert(x);
      if (x <= mx / i)
        x *= i;
      else
        break;
    }
  }
  for (auto it : all) {
    long long sq = sqrt(it);
    while (sq * sq <= it) sq++;
    while (sq * sq > it) {
      sq--;
    }
    if (sq * sq != it) a.push_back(it);
  }
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    cout << best(n) << endl;
  }
}
