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;
bool t[15000000];
bool used[15000000];
int p[15000000];
int get(int x) {
  if (p[x] == -1) return x;
  return (p[x] = get(p[x]));
}
void unite(int i1, int j1) {
  int i = get(i1);
  int j = get(j1);
  if (i != j) {
    if (i > j)
      p[i] = j;
    else
      p[j] = i;
  }
}
void add_edge(int k1, int k2) {
  if (t[k1] && t[k2]) {
    if (k1 < 10000010 && k2 < 10000010) unite(k1, k2);
  }
}
int n;
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int w = 0;
    scanf("%d", &w);
    t[w] = 1;
  }
  for (int i = 0; i < 15000000; i++) p[i] = -1;
  for (int i = 1; i * i < 15000000; i++)
    for (int j = 1; j <= i && (j * j + i * i < 15000000); j++)
      if (gcd(i, j) == 1 && (i + j) % 2 == 1) {
        int k1 = i * i - j * j;
        int k2 = 2 * i * j;
        int k3 = i * i + j * j;
        for (int k = 1;
             (k * k1) < 15000000 && (k * k2) < 15000000 && (k * k3) < 15000000;
             k++) {
          add_edge(k1, k2);
          add_edge(k2, k3);
          add_edge(k3, k1);
        }
      }
  int ans = 0;
  for (int i = 0; i < 15000000; i++)
    if (p[i] == -1 && t[i]) ans++;
  printf("%d\n", ans);
  return 0;
}