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 MX = 60 + 5, inf = 1.5e18;
long long T;
long long L[MX], H[MX];
vector<pair<long long, long long> > vec;
inline long long mul(long long a, long long b) {
  if (a > inf / b) return inf;
  return a * b;
}
inline long long pw(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
    if (res == inf) return res;
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  for (long long i = 2; i < MX; i++) {
    bool f = 0, ok = 1;
    long long x = i;
    for (long long j = 2; j <= x; j++) {
      if (x % j) continue;
      f ^= 1;
      x /= j;
      if (x % j == 0) {
        ok = 0;
        break;
      }
    }
    if (ok) {
      vec.push_back({i, (f ? -1 : 1)});
      L[i] = 1;
      long long low = 1, high = 1e9 + 5;
      while (high - low > 1) {
        long long mid = (low + high) / 2;
        if (pw(mid, i) < inf)
          low = mid;
        else
          high = mid;
      }
      H[i] = high;
    }
  }
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    long long ans = n - 1;
    for (pair<long long, long long> u : vec) {
      long long p = u.first;
      long long low = L[p], high = H[p];
      while (high - low > 1) {
        long long mid = (low + high) / 2;
        if (pw(mid, p) <= n)
          low = mid;
        else
          high = mid;
      }
      ans += (low - 1) * u.second;
    }
    cout << ans << '\n';
  }
}
