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>
const int N = 101;
long long mu[N];
long long gi(long long cur, long long many, long long x) {
  if (many == 0) return 1;
  if (many == 1) return cur;
  if (cur > x / cur) return -1;
  long long ret = gi(cur * cur, many / 2, x);
  if (ret < 0) return -1;
  if (many % 2) {
    if (ret > x / cur) return -1;
    ret *= cur;
  }
  return ret;
}
int main(void) {
  mu[1] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 2 * i; j < N; j += i) {
      mu[j] -= mu[i];
    }
  }
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    long long n;
    scanf("%lld", &n);
    int i = 1;
    long long dap = 0;
    long long bf = n;
    while ((1LL << i) <= n) {
      long long l = 2, r = bf, mx = 2;
      while (l <= r) {
        long long mid = (l + r) / 2;
        if (gi(mid, i, n) >= 0) {
          mx = mx > mid ? mx : mid;
          l = mid + 1;
        } else {
          r = mid - 1;
        }
      }
      dap += mu[i] * (mx - 1);
      bf = mx;
      i++;
    }
    printf("%lld\n", dap);
  }
}
