#include <bits/stdc++.h>
using namespace std;
int cnt, flag[105] = {1, 1}, pri[105], mu[105] = {0, 1};
long long sum[105];
void Mulset() {
  int i, j;
  for (i = 2; i <= 105; i++) {
    if (flag[i] == 0) {
      pri[++cnt] = i;
      mu[i] = -1;
    }
    for (j = 1; j <= cnt && i * pri[j] <= 105; j++) {
      flag[i * pri[j]] = 1;
      if (i % pri[j] == 0) {
        mu[i * pri[j]] = 0;
        break;
      }
      mu[i * pri[j]] = -mu[i];
    }
  }
}
int main(void) {
  long long ans, n;
  int T, i;
  Mulset();
  scanf("%d", &T);
  while (T--) {
    ans = 0;
    scanf("%lld", &n);
    for (i = 2; i <= 60; i++) {
      sum[i] = (long long)powl((long double)n + 0.1, (long double)1.0 / i) - 1;
      ans += sum[i] * mu[i];
    }
    printf("%lld\n", n - 1 + ans);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int cnt, flag[105] = {1, 1}, pri[105], mu[105] = {0, 1};
long long sum[105];
void Mulset() {
  int i, j;
  for (i = 2; i <= 105; i++) {
    if (flag[i] == 0) {
      pri[++cnt] = i;
      mu[i] = -1;
    }
    for (j = 1; j <= cnt && i * pri[j] <= 105; j++) {
      flag[i * pri[j]] = 1;
      if (i % pri[j] == 0) {
        mu[i * pri[j]] = 0;
        break;
      }
      mu[i * pri[j]] = -mu[i];
    }
  }
}
int main(void) {
  long long ans, n;
  int T, i;
  Mulset();
  scanf("%d", &T);
  while (T--) {
    ans = 0;
    scanf("%lld", &n);
    for (i = 2; i <= 60; i++) {
      sum[i] = (long long)powl((long double)n + 0.1, (long double)1.0 / i) - 1;
      ans += sum[i] * mu[i];
    }
    printf("%lld\n", n - 1 + ans);
  }
  return 0;
}

```