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];
long long gcd(long long a, long long 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() {
  while (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;
    int ymax = (int)(sqrt(Max / 2.0));
    int xmax = (int)(sqrt(Max * 3.0 / 2.0));
    for (long long y = 1; y <= ymax; y++)
      for (long long x = y + 1;
           2 * x * y <= Max && x * x - y * y <= Max && x <= xmax; x++)
        if (gcd(x * x - y * y, 2 * x * y) == 1) {
          long long a = x * x - y * y;
          long long 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[fa] = fc;
              ans--;
            }
          }
          if (pa[c] && pa[b]) {
            int fc = findset(c), fb = findset(b);
            if (fc != fb) {
              pa[fc] = fb;
              ans--;
            }
          }
        }
    cout << ans << endl;
  }
  return 0;
}