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 father[MAXN + 10];
int n;
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int find(int u) {
  int v = u, w;
  while (v != father[v]) v = father[v];
  while (father[u] != v) {
    w = father[u];
    father[u] = v;
    u = w;
  }
  return v;
}
int merge(int a, int b) {
  if (!father[a] || !father[b]) return 0;
  a = find(a);
  b = find(b);
  if (a == b) return 0;
  father[a] = b;
  return 1;
}
int main() {
  int a, i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a);
    father[a] = a;
  }
  for (i = 1; i * i <= MAXN * 4; i++)
    for (j = i + 1; j * j <= MAXN * 4; j += 2) {
      if (gcd(i, j) != 1) continue;
      int x = j * j - i * i, y = 2 * i * j, z = i * i + j * j;
      if (x <= MAXN && y <= MAXN) n -= merge(x, y);
      if (x <= MAXN && z <= MAXN) n -= merge(x, z);
      if (z <= MAXN && y <= MAXN) n -= merge(y, z);
    }
  printf("%d\n", n);
  return 0;
}