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 = (1ll << 63) - 1;
long long n;
int miu[64], v[64];
void init() {
  for (int i = 1; i < 64; i++) {
    miu[i] = 1;
    v[i] = 0;
  }
  for (int i = 2; i < 64; i++) {
    if (v[i]) continue;
    miu[i] = -1;
    for (int j = 2 * i; j < 64; j += i) {
      v[j] = 1;
      if ((j / i) % i == 0)
        miu[j] = 0;
      else
        miu[j] *= -1;
    }
  }
}
long long llpow(long long a, long long k) {
  long long res = 1;
  while (k) {
    if (a > n) return INF;
    if (k & 1) res = res * a;
    if (res > n) return INF;
    a = a * a;
    k /= 2;
  }
  return res;
}
long long get(long long n, long long k) {
  long long t = powl(n, 1. / k) - 0.5;
  return t + (powl(t + 1, k) - 0.5 <= n);
}
int main() {
  ios::sync_with_stdio(0);
  int T;
  cin >> T;
  init();
  while (T--) {
    cin >> n;
    long long ans = 0;
    for (int i = 1; i <= 62; i++) {
      if (miu[i] == 0) continue;
      ans += miu[i] * (get(n, i) - 1);
    }
    cout << ans << endl;
  }
  return 0;
}
