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;
int pn;
const int tot = 10001000;
int head[tot], tow[tot], nex[tot];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void add(int a, int b) {
  tow[pn] = a;
  nex[pn] = head[b];
  head[b] = pn++;
  tow[pn] = b;
  nex[pn] = head[a];
  head[a] = pn++;
}
void pre() {
  pn = 0;
  memset(head, -1, sizeof(head));
  int n, m, a, b, c;
  for (n = 1; n * n <= 1e7; n++) {
    for (m = n + 1; (a = 2 * m * n) <= 1e7 && (b = m * m - n * n) <= 1e7; m++) {
      if (gcd(a, b) == 1) {
        c = m * m + n * n;
        add(a, b);
        if (c <= 1e7) {
          add(b, c);
          add(a, c);
        }
      }
    }
  }
}
int n;
int x[tot];
void remv(int z) {
  x[z] = 0;
  for (int i = head[z]; i != -1; i = nex[i]) {
    if (x[tow[i]]) {
      remv(tow[i]);
    }
  }
}
int main() {
  int i, j, k;
  pre();
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &j);
    x[j] = 1;
  }
  k = 0;
  for (i = 0; i < tot; i++) {
    if (x[i]) {
      remv(i);
      k++;
    }
  }
  printf("%d\n", k);
  return 0;
}