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 int maxn = 1000000 + 10;
const int maxa = 10000000 + 10;
const int inf = (-1u) >> 1;
struct disjoint_set {
  int p[maxn];
  void clear(int sz) {
    for (int i = (1); i <= (sz); ++i) {
      p[i] = i;
    }
  }
  int find(int x) { return x == p[x] ? x : p[x] = find(p[x]); }
  bool join(int x, int y) {
    x = find(x), y = find(y);
    if (x == y) return false;
    p[x] = y;
    return true;
  }
};
int id[maxa], n;
disjoint_set ds;
bool check(int x, int y) { return id[x] && id[y] && ds.join(id[x], id[y]); }
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); ++i) {
    int t;
    scanf("%d", &t);
    id[t] = i;
  }
  ds.clear(n);
  int ans = n;
  for (int s = 1; s <= 6000; ++s) {
    for (int t = (s & 1) + 1; t < s; t += 2) {
      if (gcd(s, t) != 1) continue;
      int x = 2 * s * t;
      int y = s * s - t * t;
      int z = s * s + t * t;
      if (x >= maxa || y >= maxa) {
        continue;
      }
      ans -= check(x, y);
      if (z < maxa) {
        ans -= check(x, z);
        ans -= check(y, z);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}