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 pre[1000003];
int a[1000003];
int v[10000010];
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int ufind(int u) {
  if (pre[u] == u) return u;
  return pre[u] = ufind(pre[u]);
}
void dounion(int u, int v) {
  u = ufind(u), v = ufind(v);
  pre[u] = v;
}
int main() {
  int N;
  scanf("%d", &N);
  for (int i = (0); i < (int)(N); i++) scanf("%d", a + i);
  for (int i = (0); i < (int)(10000010); i++) v[i] = -1;
  for (int i = (0); i < (int)(N); i++) v[a[i]] = i;
  for (int i = (0); i < (int)(N); i++) pre[i] = i;
  for (long long i = 1; i * i < 10000010; i++)
    for (long long j = i + 1; max(j * j - i * i, 2 * j * i) <= 10000010;
         j += 2) {
      if (gcd(i, j) == 1) {
        long long a = 2 * i * j, b = j * j - i * i, c = i * i + j * j;
        if (a < 10000010 && b < 10000010 && v[a] != -1 && v[b] != -1)
          dounion(v[a], v[b]);
        if (a < 10000010 && c < 10000010 && v[a] != -1 && v[c] != -1)
          dounion(v[a], v[c]);
        if (b < 10000010 && c < 10000010 && v[b] != -1 && v[c] != -1)
          dounion(v[b], v[c]);
      }
    }
  int ret = 0;
  for (int i = (0); i < (int)(N); i++)
    if (pre[i] == i) ret++;
  printf("%d\n", ret);
  return 0;
}