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 = 100;
int prim[N], cnt;
int mbus[N], vis[N];
void get_mbus() {
  memset(vis, 0, sizeof(vis));
  memset(mbus, 0, sizeof(mbus));
  vis[0] = vis[1] = 1;
  mbus[1] = 1;
  cnt = 0;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      prim[cnt++] = i;
      mbus[i] = -1;
    }
    for (int j = 0; i * prim[j] < N && j < cnt; j++) {
      vis[i * prim[j]] = 1;
      if (i % prim[j] == 0) {
        mbus[i * prim[j]] = 0;
        break;
      } else {
        mbus[i * prim[j]] = -1 * mbus[i];
      }
    }
  }
}
long long gen(long long n, long long k) {
  long long t = powl(n, 1.0 / k) - 0.5;
  return t + (powl(t + 1, k) - 0.5 <= n);
}
int main() {
  get_mbus();
  int T;
  long long n;
  scanf("%d", &T);
  while (T--) {
    scanf("%lld", &n);
    long long ans = n - 1;
    for (int i = 2; i < 100; i++) {
      if (mbus[i] == 0) continue;
      long long x = gen(n, i);
      if (x == 1) break;
      ans += mbus[i] * (x - 1);
    }
    printf("%lld\n", ans);
  }
  return 0;
}
