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;
int lim = 0, n, res;
long long x, y, z;
int a[2000005], f[2000005 * 10];
long long gcd(long long x, long long y) {
  if (!y) return x;
  return gcd(y, x % y);
}
int getfa(int x) {
  int y, old;
  for (y = x; y != f[y]; y = f[y])
    ;
  for (; x != f[x]; x = old) {
    old = f[x];
    f[x] = y;
  }
  return y;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    lim = max(lim, a[i]);
    f[a[i]] = a[i];
  }
  res = n;
  for (int i = 2; i + i <= lim; ++i)
    for (int j = 1; j < i; ++j) {
      x = (long long)i * i - (long long)j * j;
      y = 2LL * i * j;
      if (y > lim) break;
      if (x <= lim && (f[x] || f[y]) && (x & 1) && gcd(x, y) == 1) {
        if (f[x] && f[y] && getfa(x) != getfa(y)) f[getfa(x)] = getfa(y), --res;
        z = (long long)i * i + (long long)j * j;
        if (z <= lim && f[z]) {
          if (f[x] && getfa(z) != getfa(x)) f[getfa(z)] = getfa(x), --res;
          if (f[y] && getfa(z) != getfa(y)) f[getfa(z)] = getfa(y), --res;
        }
      }
    }
  printf("%d\n", res);
  return 0;
}