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;
using ll = long long;
vector<ll> primes({2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53,
                   59});
vector<pair<ll, ll>> powers;
void GenPowers() {
  for (int mask = 1; mask < (1 << primes.size()); mask++) {
    ll value = 1;
    for (int i = 0; i < primes.size(); i++) {
      if (mask & (1 << i)) {
        value *= primes[i];
        if (value > 59) {
          break;
        }
      }
    }
    if (value < 60) {
      if (__builtin_popcount(mask) & 1) {
        powers.push_back({value, 1});
      } else {
        powers.push_back({value, -1});
      }
    }
  }
  sort(powers.begin(), powers.end());
}
bool check(ll n, ll p, ll x) {
  ll val = 1ll;
  for (ll i = 0; i < p; i++) {
    val *= x;
    if (val > n) return false;
  }
  return true;
}
ll count(ll n, ll p) {
  ll x = floor(exp(log(n) / p) + 0.1);
  while (not check(n, p, x)) x--;
  return x - 1ll;
}
ll query(ll n) {
  ll res = n - 1;
  for (auto p : powers) {
    res -= p.second * count(n, p.first);
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  GenPowers();
  int t;
  cin >> t;
  for (int tc = 0; tc < t; tc++) {
    ll n;
    cin >> n;
    cout << query(n) << '\n';
  }
}
