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 = 10000000;
int pos[maxn + 10], a[1001000];
int fa[1001000], n, ans;
void init() {
  for (int i = 1; i <= n; i++) fa[i] = i;
}
int getfa(int v) { return fa[v] == v ? v : fa[v] = getfa(fa[v]); }
void merge(int x, int y) {
  if (x == 0 || y == 0) return;
  int p = getfa(x), q = getfa(y);
  if (p != q) ans--, fa[q] = p;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  init();
  ans = n;
  for (int i = 1; i <= 5050; i++)
    for (int j = i % 2 + 1; j < i; j += 2)
      if (gcd(i, j) == 1) {
        int x = i * i - j * j;
        int y = 2 * i * j;
        if (x > maxn || y > maxn) continue;
        merge(pos[x], pos[y]);
        int z = i * i + j * j;
        if (z <= maxn) {
          merge(pos[y], pos[z]);
          merge(pos[x], pos[z]);
        }
      }
  printf("%d\n", ans);
  return 0;
}