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;
using ll = long long;
using ld = long double;
const int N = 100 + 7;
ll n, m, k;
int t;
int primes[N], mu[N], cnt;
bool vis[N];
void init(int n) {
  vis[1] = 1;
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (vis[i] == 0) {
      primes[++cnt] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= cnt && i * primes[j] <= n; ++j) {
      vis[i * primes[j]] = 1;
      if (i % primes[j] == 0) {
        mu[i * primes[j]] = 0;
        break;
      }
      mu[i * primes[j]] -= mu[i];
    }
  }
}
void solve() {
  scanf("%lld", &n);
  ll ans = 0;
  for (int i = 2; i <= 60; ++i) {
    if (mu[i] == 0) continue;
    ll res = (ll)(pow(n, 1.0 / i)) + 2;
    while (pow(res, i) > n) res--;
    ans += (ll)mu[i] * (res - 1);
  }
  printf("%lld\n", n - 1 + ans);
}
int main() {
  init(N - 7);
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}
