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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int n;
int m, r = 0;
int z[10000001];
int a[1000001];
int p[1000001];
int l;
int ans = 0;
int getp(int x) {
  if (p[x] != x) p[x] = getp(p[x]);
  return p[x];
}
int un(int a, int b) {
  int pa = getp(a);
  int push_back = getp(b);
  if (pa != push_back) {
    ans--;
    if (rand() & 1)
      p[pa] = push_back;
    else
      p[push_back] = pa;
  }
  return 0;
}
int gcd(int a, int b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int main() {
  scanf("%d", &n);
  memset(z, -1, sizeof(z));
  for (int i = 0; i < n; i++) {
    p[i] = i;
    scanf("%d", &a[i]);
    z[a[i]] = i;
  }
  ans = n;
  int sum = 0;
  for (int x = 2; x <= 4000; x++)
    for (int y = 1; y < x; y++)
      if (((x ^ y) & 1) && gcd(x, y) == 1) {
        long long a = (long long)x * x - (long long)y * y;
        long long b = (long long)2 * x * y;
        long long c = (long long)x * x + (long long)y * y;
        if (a > b) swap(a, b);
        if (b > c) swap(b, c);
        if (a > b) swap(a, b);
        if (b <= 10000000 && z[a] != -1 && z[b] != -1) un(z[a], z[b]);
        if (c <= 10000000 && z[a] != -1 && z[c] != -1) un(z[a], z[c]);
        if (c <= 10000000 && z[c] != -1 && z[b] != -1) un(z[c], z[b]);
      }
  cerr << sum << endl;
  printf("%d\n", ans);
  return 0;
}