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>
int fa[10000000 + 10];
inline long long gcd(long long a, long long b) {
  while (a > 0) {
    long long tmp = a;
    a = b % a;
    b = tmp;
  }
  return b;
}
int getfa(int a) {
  int f = a, w;
  while (fa[f] != f) f = fa[f];
  while (fa[a] != a) {
    w = fa[a];
    fa[a] = f;
    a = w;
  }
  return f;
}
int m(long long a, long long b) {
  if (fa[a] == 0 || fa[b] == 0) return 0;
  int faa = getfa(a), fbb = getfa(b);
  if (faa == fbb) return 0;
  fa[faa] = fbb;
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int tmp;
    scanf("%d", &tmp);
    fa[tmp] = tmp;
  }
  for (long long i = 2; i * i <= 2 * 10000000; i += 2)
    for (long long j = i + 1, a = i / 2 * (j + 1), b = j, c = i / 2 * i + j;
         a <= 10000000 && b <= 10000000; j += 2, a += i, b += j, c += j) {
      if (gcd(a, b) > 1) continue;
      n -= m(a, b);
      if (c <= 10000000) n -= m(b, c) + m(c, a);
    }
  printf("%d\n", n);
}