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 N = 1e18;
void mul(long long &x, long long y) {
  if (N / y < x)
    x = N + 1;
  else
    x *= y;
}
long long binpow(long long u, int v) {
  long long ans = 1;
  while (v) {
    if (v & 1) mul(ans, u);
    mul(u, u);
    v >>= 1;
  }
  return ans;
}
long long p[70][70];
int main() {
  int T;
  scanf("%d", &T);
  for (int i = 1; i < 70; ++i) {
    p[i][0] = 1;
    for (int j = 1; j < 70; ++j) {
      p[i][j] = p[i][j - 1];
      mul(p[i][j], i);
    }
  }
  while (T--) {
    long long n;
    scanf("%lld", &n);
    vector<long long> f(61, 0);
    for (int i = 60; i >= 1; --i) {
      long long lo = 1, hi = n;
      if (i >= 10) {
        int x = 1;
        while (p[x + 1][i] <= n) ++x;
        f[i] = x - 1;
      } else
        while (lo <= hi) {
          long long m = (lo + hi) >> 1;
          long long v = 1;
          for (int k = 1; k <= i; ++k) mul(v, m);
          if (v <= n) {
            f[i] = m - 1;
            lo = m + 1;
          } else
            hi = m - 1;
        }
      for (int j = i + i; j <= 60; j += i) {
        f[i] -= f[j];
      }
    }
    cout << f[1] << endl;
  }
  return 0;
}
