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 primes[17] = {2,  3,  5,  7,  11, 13, 17, 19, 23,
                        29, 31, 37, 41, 43, 47, 53, 59};
int T;
long long N;
vector<pair<int, int> > E;
long long expo(long long num, int p) {
  long long ret = 1;
  for (int i = 0; i < p; i++) {
    if (ret > N / num) return N + 1;
    ret *= num;
  }
  return ret;
}
long long get_root(long long hi, int p) {
  long long lo = 1;
  while (lo < hi) {
    long long mid = lo + (hi - lo + 1) / 2;
    if (expo(mid, p) <= N)
      lo = mid;
    else
      hi = mid - 1;
  }
  return lo;
}
int main() {
  scanf("%d", &T);
  for (int mask = 1; mask < (1 << 17); mask++) {
    int s = 1, e = 1;
    bool good = true;
    for (int i = 0; i < 17; i++)
      if (mask & (1 << i)) {
        e *= primes[i];
        if (e >= 60) {
          good = false;
          break;
        }
        s *= -1;
      }
    if (good) E.push_back(make_pair(e, s));
  }
  sort(E.begin(), E.end());
  for (int t = 0; t < T; t++) {
    scanf("%lld", &N);
    long long ans = N - 1, hi = 1000000001;
    for (int i = 0; i < E.size(); i++) {
      hi = get_root(hi, E[i].first);
      ans += E[i].second * (hi - 1);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
