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;
char ch;
const int mn = 10000024;
int n, i, a, ans, fa[mn];
void get(int &a) {
  int p = 1;
  for (ch = getchar(); (ch < '0' || ch > '9') && (ch != '-'); ch = getchar())
    ;
  if (ch == '-')
    a = 0, p = -1;
  else
    a = ch - '0';
  for (ch = getchar(); ch >= '0' && ch <= '9'; ch = getchar())
    a = a * 10 + ch - '0';
  a *= p;
}
int find(int r) { return fa[r] == -1 ? r : fa[r] = find(fa[r]); }
void merge(int a, int b) {
  if (a < mn && b < mn && fa[a] && fa[b] && find(a) != find(b))
    fa[find(a)] = find(b), ans--;
}
void gen(int a, int b, int c) {
  if (a >= mn || b >= mn) return;
  merge(a, b), merge(a, c), merge(b, c);
  gen(a - 2 * b + 2 * c, 2 * a - b + 2 * c, 2 * a - 2 * b + 3 * c);
  gen(a + 2 * b + 2 * c, 2 * a + b + 2 * c, 2 * a + 2 * b + 3 * c);
  gen(-a + 2 * b + 2 * c, -2 * a + b + 2 * c, -2 * a + 2 * b + 3 * c);
}
int main() {
  get(n);
  for (i = 1; i <= n; ++i) {
    get(a);
    fa[a] = -1;
  }
  ans = n;
  gen(3, 4, 5);
  printf("%d\n", ans);
  cin.get();
  cin.get();
  return 0;
}