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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T, typename U>
U type_cast(const T& a) {
  U res;
  stringstream ss;
  ss << a;
  ss >> res;
  return res;
}
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T> >;
template <typename T>
using max_queue = priority_queue<T>;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
const int inf = 0x3f3f3f3f;
const int Maxn = 300005;
ll fastpow(ll a, ll b) {
  ll res = 1LL;
  for (; b; b >>= 1, a *= a)
    if (b & 1) res *= a;
  return res;
}
int q;
ll n;
vector<ll> v;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  for (ll i = 2; i <= 1000000; ++i) {
    ll x = i * i;
    for (;;) {
      if (1000000000000000000 / x >= i) {
        x *= i;
        ll p = (ll)sqrt(x);
        if (p * p == x) continue;
        v.push_back(x);
      } else
        break;
    }
  }
  sort(v.begin(), v.end()), v.erase(unique(v.begin(), v.end()), v.end());
  for (cin >> q; q--;) {
    cin >> n;
    cout << n - 1 - (upper_bound(v.begin(), v.end(), n) - v.begin()) -
                (ll(sqrt(n)) - 1)
         << endl;
  }
  return 0;
}
