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 BORD = 1e18;
int q;
long long n;
long long bates[60];
long long fpow(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = fpow(a, b / 2);
  temp = temp * temp;
  if (b & 1) temp = temp * a;
  return temp;
}
long double ffpow(long double a, long long b) {
  if (b == 0) return 1.0;
  long double temp = ffpow(a, b / 2);
  temp = temp * temp;
  if (b & 1) temp = temp * a;
  return temp;
}
long long calc(long long n, long long p) {
  long long low = 1, high = bates[p] - 1;
  while (low < high) {
    long long mid = (low + high + 1) / 2;
    if (fpow(mid, p) <= n) {
      low = mid;
    } else {
      high = mid - 1;
    }
  }
  return low - 1LL;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  bates[2] = 1e9 + 1;
  long long tempid = 59;
  for (long long i = 2; tempid > 2; i++) {
    long double temp = ffpow((long double)i, tempid);
    while (temp > 1e18) {
      bates[tempid] = i;
      tempid--;
      temp = ffpow((long double)i, tempid);
    }
  }
  cin >> q;
  while (q--) {
    cin >> n;
    long long ans = 0;
    long long count[60];
    fill(count, count + 60, 1);
    for (long long i = 2; i < 60; i++) {
      ans += count[i] * calc(n, i);
      for (long long j = i + i; j < 60; j += i) {
        count[j] -= count[i];
      }
    }
    cout << n - ans - 1LL << '\n';
  }
  return 0;
}
