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;
void Freopen() {
  freopen(
      "title"
      ".in",
      "r", stdin);
  freopen(
      "title"
      ".out",
      "w", stdout);
}
long long read() {
  long long g = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    g = g * 10 + ch - '0';
    ch = getchar();
  }
  return g * f;
}
long long n;
int can[100], nu[100];
long double ksm(long double x, int y) {
  long double re = 1;
  while (y) {
    if (y & 1) re *= x;
    x *= x;
    y >>= 1;
  }
  return re;
}
void solve() {
  n = read();
  long long ans = n - 1;
  for (int i = (2); i <= (63); i++)
    if (!can[i]) {
      int re = (int)pow(n, 1. / i) + 2;
      while (ksm(re, i) > n) re--;
      ans -= ((nu[i] & 1) ? 1 : -1) * (re - 1);
    }
  cout << ans << '\n';
}
signed main() {
  for (int i = (2); i <= (63); i++) {
    int now = i;
    for (int j = (2); j <= (63); j++)
      if (!(now % j)) {
        int re = 0;
        while (!(now % j)) re++, now /= j;
        if (re > 1) can[i] = 1;
        nu[i]++;
      }
  }
  for (int T = read(); T--;) solve();
  return signed();
}
