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;
long long n, R, RP[70];
int T, nm = 1, o;
int GCD(int a, int b) { return !a ? b : GCD(b % a, a); }
void BT(int a) {
  if (a == 62) {
    if (o == 0) return;
    if (o % 2 == 0)
      RP[nm]--;
    else
      RP[nm]++;
    return;
  }
  BT(a + 1);
  int ls = nm;
  nm = (nm * a) / GCD(nm, a);
  o++;
  if (nm <= 62) BT(a + 1);
  nm = ls;
  o--;
}
long long powe(long long a, int b) {
  long long res = 1;
  for (; b; b /= 2) {
    if (b % 2) res *= a;
    a *= a;
  }
  return res;
}
long long gen(long long k) {
  long long t = powl(n, 1. / k) - 0.5;
  return t + (powl(t + 1, k) - 0.5 <= n);
}
int main() {
  cin >> T;
  BT(2);
  for (int i = 0; i < T; i++) {
    cin >> n;
    R = 0;
    for (int j = 2; j <= 62; j++) R += RP[j] * (gen(j) - 1LL);
    cout << n - R - 1 << endl;
  }
}
