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;
template <typename T>
void read(T &t) {
  t = 0;
  char ch = getchar();
  int f = 1;
  while ('0' > ch || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  do {
    (t *= 10) += ch - '0';
    ch = getchar();
  } while ('0' <= ch && ch <= '9');
  t *= f;
}
int T, p[19] = {0,  2,  3,  5,  7,  11, 13, 17, 19, 23,
                29, 31, 37, 41, 43, 47, 53, 59, 61};
int v[40][4], tot;
long long q[40];
long long n, ans;
long long gcd(long long x, long long y) {
  if (!x || !y) return x + y;
  return gcd(y, x % y);
}
long long gen(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() {
  for (int t = 1; t < (1 << 18); t++) {
    long long tmp = 1;
    for (int i = 1; i <= 18; i++)
      if (t & (1 << (i - 1))) {
        tmp = tmp / gcd(tmp, p[i]) * p[i];
        if (tmp > 63) break;
      }
    if (tmp <= 63) {
      tot++;
      q[tot] = tmp;
      for (int i = 1; i <= 18; i++)
        if (t & (1 << (i - 1))) v[tot][++v[tot][0]] = p[i];
    }
  }
  read(T);
  while (T--) {
    read(n);
    ans = 0;
    long long tmp, t2;
    for (int t = 1; t <= tot; t++) {
      if (v[t][0] % 2 == 0)
        tmp = -1;
      else
        tmp = 1;
      t2 = gen(n, q[t]);
      ans += tmp * t2;
    }
    printf("%lld\n", n + 1 - ans);
  }
  return 0;
}
