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;
int t, mc[60];
long long n, rb;
long long a(int p) {
  long long lb = 1;
  while (lb < rb) {
    long long mb = (lb + rb + 1) / 2, mp = 1;
    for (int i = 0; i < p; ++i, mp *= mb) {
      if (n / mb < mp) {
        mp = n + 1;
        break;
      }
    }
    if (mp <= n)
      lb = mb;
    else
      rb = mb - 1;
  }
  return rb - 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  for (int i = 1; i < 60; ++i) {
    int tc = i == 1;
    mc[i] = tc - mc[i];
    for (int j = i; (j += i) < 60;) mc[j] += mc[i];
  }
  cin >> t;
  while (t--) {
    cin >> n;
    long long ans = n - 1;
    rb = (long long)1e9;
    for (int i = 2; i < 60; ++i)
      if (mc[i]) ans += mc[i] * a(i);
    cout << ans << "\n";
  }
}
