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;
inline long long binpow(long long x, long long deg) {
  if (pow(x, deg) > 2e18) return 2e18;
  long long ans = 1;
  while (deg > 0) {
    if (deg & 1) ans *= x;
    deg >>= 1;
    x *= x;
  }
  return ans;
}
inline long long root(long long x, long long deg) {
  long long ans = pow(x, 1.0 / deg);
  while (binpow(ans, deg) < x) ++ans;
  while (binpow(ans, deg) > x) --ans;
  return ans;
}
int t, mobius[60];
long long x;
int main() {
  ios_base::sync_with_stdio(false);
  mobius[1] = 1;
  for (int i = 2; i <= 59; ++i) {
    int sq = 0;
    int cnt = 0;
    int x = i;
    for (int j = 2; j * j <= x; ++j) {
      if (x % j == 0) {
        ++cnt;
        int cur = 0;
        while (x % j == 0) x /= j, ++cur;
        if (cur > 1) sq = 1;
      }
    }
    if (x > 1) ++cnt;
    mobius[i] = (sq ? 0 : cnt % 2 ? -1 : 1);
  }
  cin >> t;
  while (t--) {
    cin >> x;
    long long ans = x - 1;
    for (int i = 2; i <= 59 && (1LL << i) <= x; ++i)
      ans += mobius[i] * (root(x, i) - 1);
    cout << ans << "\n";
  }
}
