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 long long int INF = 2e18;
vector<int> v, d;
double bp(double x, int p) {
  double r = 1;
  while (p > 0) {
    if (p & 1) {
      r *= x;
    }
    p >>= 1;
    x = x * x;
  }
  return r;
}
long long int getRoot(long long int x, int p) {
  long long int l = 0, r = p > 8 ? 200 : 2e9;
  while (r - l > 1) {
    long long int m = (r + l) >> 1;
    double t = bp(m, p);
    if (t < INF && (long long int)t <= x)
      l = m;
    else {
      r = m;
    }
  }
  return l;
}
int main() {
  for (int i = 2; i <= 64; i++) {
    bool ok = true;
    int t = i, c = 0;
    for (int j = 2; j <= t; j++) {
      int q = 0;
      while (t % j == 0) {
        t /= j;
        q++;
      }
      if (q > 1) {
        ok = false;
        break;
      }
      if (q) c ^= 1;
    }
    if (ok) {
      v.push_back(i);
      d.push_back(c ? -1 : 1);
    }
  }
  int t;
  scanf("%d", &t);
  while (t--) {
    long long int n;
    scanf("%lld", &n);
    long long int ans = n - 1;
    for (int i = 0; i < v.size(); i++) {
      long long int t = getRoot(n, v[i]) - 1;
      ans += t * d[i];
    }
    printf("%lld\n", ans);
  }
  return 0;
}
