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 N = 1e6 + 6, P = 60;
const long long M = 1e18 + 18;
int t;
long long m, cnt[N];
vector<long long> powers[P];
long long get_pow(int x, int pw) {
  long long cur = 1;
  for (int j = 0; j < pw; ++j) {
    if (x > M / cur) return 0;
    cur *= x;
  }
  return cur;
}
void prepare() {
  for (int i = 2; i < P; ++i) {
    for (int x = 2; x < N; ++x) {
      long long cur = get_pow(x, i);
      if (!cur) break;
      powers[i].push_back(cur);
    }
  }
}
long long get_count(long long x, long long n) {
  if (x == 2) return sqrt(n) - 1;
  auto &q = powers[x];
  return upper_bound(q.begin(), q.end(), n) - q.begin();
}
long long get_correct_count(long long x) {
  for (long long i = x + x; i <= m; i += x) cnt[x] -= cnt[i];
  return cnt[x];
}
void solve() {
  long long n;
  scanf("%lld", &n);
  m = 0;
  for (int i = 2;; ++i) {
    long long cur = get_count(i, n);
    if (!cur) break;
    m = i;
    cnt[i] = cur;
  }
  long long ans = n - 1;
  for (int i = m; i > 1; --i) ans -= get_correct_count(i);
  printf("%lld\n", ans);
}
int main() {
  prepare();
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) solve();
  return 0;
}
