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 int MAXN = 1000010;
const int MAX = 10000000;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
class DSet {
 public:
  int p[MAXN], t;
  void init() { memset(p, 0, sizeof(p)); }
  void setFriend(int i, int j) {
    for (; p[t = i]; i = p[i], p[t] = (p[i] ? p[i] : i))
      ;
    for (; p[t = j]; j = p[j], p[t] = (p[j] ? p[j] : j))
      ;
    p[i] = (i == j ? 0 : j);
  }
  bool isFriend(int i, int j) {
    for (; p[t = i]; i = p[i], p[t] = (p[i] ? p[i] : i))
      ;
    for (; p[t = j]; j = p[j], p[t] = (p[j] ? p[j] : j))
      ;
    return i == j && i;
  }
} p;
int mp[10000010];
int main() {
  int i, j, k, n, m, mx, my, a, b, c, x, y;
  m = 0;
  while (scanf("%d", &n) != EOF) {
    for (i = 1; i <= n; i++) {
      scanf("%d", &k);
      mp[k] = i;
      if (k > m) m = k;
    }
    mx = sqrt(3 * m / 2.0);
    my = sqrt(m / 2.0);
    p.init();
    for (y = 1; y <= my; y++)
      for (x = y; x <= mx; x++)
        if (gcd(x, y) == 1 && x % 2 != y % 2) {
          a = x * x - y * y, b = 2 * x * y, c = x * x + y * y;
          if (gcd(a, b) > 1 || gcd(b, c) > 1 || gcd(a, c) > 1) continue;
          if (a <= MAX && b <= MAX && mp[a] && mp[b]) {
            p.setFriend(mp[a], mp[b]);
          }
          if (a <= MAX && c <= MAX && mp[a] && mp[c]) {
            p.setFriend(mp[a], mp[c]);
          }
          if (b <= MAX && c <= MAX && mp[b] && mp[c]) {
            p.setFriend(mp[b], mp[c]);
          }
        }
    k = 0;
    for (i = 1; i <= n; i++)
      if (p.p[i] == 0) k++;
    printf("%d\n", k);
  }
}