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;
const int MAXN = 100000;
const long long INF = 1e18;
long long mul(long long a, long long b) {
  if (b == 0) return 0;
  if (a > INF / b) return INF + 9487;
  return a * b;
}
long long qpow(long long a, long long p) {
  long long res = 1;
  while (p) {
    if (p & 1) res = mul(res, a);
    a = mul(a, a);
    p >>= 1;
  }
  return res;
}
bool I_am_not_a_square_number(long long x) {
  long long sq = sqrt(x);
  for (int i = -1; i <= 1; i++)
    if ((sq + i) * (sq + i) == x) return false;
  return true;
}
vector<long long> never_elegant;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  for (long long i = 2; i <= 1000000; i++) {
    long long tmp = i * i * i;
    while (tmp <= INF) {
      if (I_am_not_a_square_number(tmp)) {
        never_elegant.push_back(tmp);
      }
      tmp = mul(tmp, i);
    }
  }
  sort(never_elegant.begin(), never_elegant.end());
  never_elegant.resize(unique(never_elegant.begin(), never_elegant.end()) -
                       never_elegant.begin());
  int T;
  cin >> T;
  while (T--) {
    long long n, sq;
    cin >> n;
    sq = sqrt(n);
    for (long long i = max(0LL, sq - 3);; i++) {
      if (i * i > n) {
        sq = i - 1;
        break;
      }
    }
    long long ans = max(0LL, sq - 1) + (upper_bound(never_elegant.begin(),
                                                    never_elegant.end(), n) -
                                        never_elegant.begin());
    cout << n - 1 - ans << "\n";
  }
}
