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 + 1;
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2) {
      if (double(res) > 1e18 / double(a)) {
        return 2e18;
      }
      res = 1LL * res * a;
    }
    if (double(a) > 1e18 / double(a)) {
      return (b == 1 ? res : 2e18);
    }
    a = 1LL * a * a;
    b /= 2;
  }
  return res;
}
long long find(long long x, long long p, long long range) {
  long long low = 1, high = range, ans = 1;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (power(mid, p) <= x) {
      ans = mid;
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  return ans;
}
int main() {
  int test;
  long long n, root[70];
  scanf("%d", &test);
  while (test--) {
    scanf("%lld", &n);
    root[1] = n - 1;
    int cnt = 1;
    for (int i = 2; i <= 60; i++) {
      long long tot = find(n, i, root[i - 1] + 1);
      root[i] = tot - 1;
      cnt = i;
    }
    for (int i = cnt; i >= 1; i--) {
      for (int j = i + i; j <= cnt; j += i) {
        root[i] -= root[j];
      }
    }
    printf("%lld\n", root[1]);
  }
  return 0;
}
