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 n, m, k;
bool debug = false;
int ans, X, Y;
long long gcd(long long x, long long y) {
  long long z = y;
  while (x % y != 0) {
    z = x % y;
    x = y;
    y = z;
  }
  return z;
}
long long a[10000000];
long long ff(long long x) {
  while (a[x] != x) x = a[x];
  return x;
}
void dd(long long x, long long y) {
  if (x > 10000000 || y > 10000000) return;
  if (a[x] != 0 && a[y] != 0) {
    long long fx = ff(x);
    long long fy = ff(y);
    if (fx != fy) {
      ans--;
      a[fx] = fy;
    }
  }
}
int main() {
  scanf("%d", &n);
  int mx = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", &m);
    a[m] = m;
    mx = max(mx, m);
  }
  X = (int)sqrt((double)mx * 3.0 / 2.0) + 1;
  Y = (int)sqrt((double)mx / 2.0) + 1;
  ans = n;
  for (long long i = 1; i <= Y; i++) {
    for (long long j = i + 1; j <= X; j += 2) {
      if (gcd(i, j) == 1) {
        long long x = j * j - i * i, y = 2 * i * j, z = i * i + j * j;
        dd(x, y);
        dd(y, z);
        dd(x, z);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}