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;
const long long MAX = 10000001;
int N;
int a[MAX], f[MAX];
int ans;
long long gcd(long long a, long long b) {
  if (b == 0LL) return a;
  return gcd(b, a % b);
}
int find(int x) {
  if (f[x] == x) return x;
  return f[x] = find(f[x]);
}
void merge(int x, int y) {
  if (!f[x] || !f[y]) return;
  int rx = find(x), ry = find(y);
  if (rx != ry) {
    ans--;
    f[rx] = ry;
  }
}
int main() {
  int i, j, k;
  cin >> N;
  ans = N;
  for (i = 0; i < N; ++i) {
    cin >> a[i];
    f[a[i]] = a[i];
  }
  for (long long i = 1; i < MAX; ++i)
    for (long long j = i + 1; 2 * i * j < MAX && j * j - i * i < MAX; j += 2) {
      if (gcd(i, j) != 1) continue;
      merge(j * j - i * i, 2 * i * j);
      if (i * i + j * j < MAX) {
        merge(i * i + j * j, j * j - i * i);
        merge(i * i + j * j, 2 * i * j);
      }
    }
  cout << ans << endl;
  return 0;
}