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 Max = 10000000;
int n;
int a[1000005];
int pa[10000005];
bool vis[10000005];
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int findset(int x) { return pa[x] == x ? x : (pa[x] = findset(pa[x])); }
int main() {
  cin >> n;
  memset(pa, 0, sizeof(pa));
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    pa[a[i]] = a[i];
  }
  int ans = n;
  for (long long x = 1; x <= (int)(sqrt(Max / 2.0 * 3.0)); x++)
    for (long long y = 1; y < x; y++)
      if (x * x - y * y <= Max && 2 * x * y <= Max &&
          gcd(x * x - y * y, 2 * x * y) == 1) {
        int a = x * x - y * y;
        int b = 2 * x * y;
        long long c = x * x + y * y;
        if (pa[a] && pa[b]) {
          int fa = findset(a), fb = findset(b);
          if (fa != fb) {
            pa[fb] = fa;
            ans--;
          }
        }
        if (c > Max) continue;
        if (pa[a] && pa[c]) {
          int fa = findset(a), fc = findset(c);
          if (fa != fc) {
            pa[fc] = fa;
            ans--;
          }
        }
        if (pa[c] && pa[b]) {
          int fc = findset(c), fb = findset(b);
          if (fc != fb) {
            pa[fb] = fc;
            ans--;
          }
        }
      }
  cout << ans << endl;
  return 0;
}