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 INF = 1e18;
int T;
vector<int> prime;
vector<long long> p[65];
int vis[105], mu[105];
long long n;
void seive(int n) {
  mu[1] = vis[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!vis[i]) prime.push_back(i), mu[i] = -1;
    for (int j = 0; j < prime.size(); ++j) {
      if (prime[j] * i > n) break;
      vis[prime[j] * i] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[prime[j] * i] = -mu[i];
    }
  }
}
void init() {
  seive(100);
  for (long long i = 1; i <= 1000000LL; ++i) {
    long long cur = i * i * i;
    for (int j = 3; j <= 62; ++j) {
      p[j].push_back(cur);
      if (cur > INF / i) break;
      cur *= i;
    }
  }
}
int main() {
  init();
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    long long x = sqrt(n + 0.5);
    long long ans = mu[2] * (x - 1);
    for (int i = 3; i <= 60; ++i) {
      long long k = upper_bound(p[i].begin(), p[i].end(), n) - p[i].begin() - 1;
      ans += k * mu[i];
    }
    printf("%lld\n", ans + n - 1);
  }
}
