Problem: Misha decided to help Pasha and Akim be friends again. He had a cunning plan — to destroy all the laughy mushrooms. He knows that the laughy mushrooms can easily burst when they laugh. Mushrooms grow on the lawns. There are a[t] mushrooms on the t-th lawn.

Misha knows that the lawns where the mushrooms grow have a unique ability. A lawn (say, i) can transfer laugh to other lawn (say, j) if there exists an integer (say, b) such, that some permutation of numbers a[i], a[j] and b is a beautiful triple (i ≠ j). A beautiful triple is such three pairwise coprime numbers x, y, z, which satisfy the following condition: x2 + y2 = z2.

Misha wants to know on which minimal number of lawns he should laugh for all the laughy mushrooms to burst.

Input

The first line contains one integer n (1 ≤ n ≤ 106) which is the number of lawns. The next line contains n integers ai which are the number of mushrooms on the i-lawn (1 ≤ ai ≤ 107). All the numbers are different.

Output

Print a single number — the minimal number of lawns on which Misha should laugh for all the mushrooms to burst.

Examples

Input

1
2


Output

1


Input

2
1 2


Output

2


Input

2
3 5


Output

1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long maxn = 10000001;
long f[maxn], n, maxN;
long find(long x) {
  if (f[x] == x)
    return x;
  else {
    f[x] = find(f[x]);
    return f[x];
  }
}
void get(long x, long y) {
  x = find(x);
  y = find(y);
  if (x != y) {
    f[y] = x;
  }
}
long gcd(long x, long y) {
  while (y > 0) {
    long t = y;
    y = x % y;
    x = t;
  }
  return x;
}
int main() {
  memset(f, 0, sizeof(f));
  cin >> n;
  for (long i = 1; i <= n; i++) {
    long x;
    cin >> x;
    f[x] = x;
    if (x > maxN) {
      maxN = x;
    }
  }
  for (long x = 1; x <= ceil(sqrt(double(maxN) / 2)); x++)
    for (long y = x + 1; y <= ceil(sqrt(double(maxN) * 3 / 2)); y++)
      if (((x + y) % 2 == 1) && (gcd(x, y) == 1)) {
        long a = y * y - x * x, b = 2 * x * y, c = x * x + y * y;
        if (a <= maxN && b <= maxN && f[a] && f[b]) get(a, b);
        if (a <= maxN && c <= maxN && f[a] && f[c]) get(a, c);
        if (b <= maxN && c <= maxN && f[b] && f[c]) get(b, c);
      }
  long ans = 0;
  for (long i = 1; i <= maxN; i++)
    if ((f[i]) && (find(i) == i)) ans++;
  cout << ans << endl;
  return 0;
}