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>
const int MAXN = 1 << 20;
const int RLIMIT = 5050;
const int LIMIT = 10000000;
struct DisjointSet {
  int p[MAXN];
  void init(int n) {
    for (int i = 1; i <= n; ++i) {
      p[i] = i;
    }
  }
  int root(int i) { return p[i] == i ? i : p[i] = root(p[i]); }
  bool link(int i, int j) {
    i = root(i);
    j = root(j);
    if (i == j) {
      return false;
    } else {
      p[i] = j;
      return true;
    }
  }
} ds;
int id[LIMIT];
inline int gao(int i, int j) {
  return id[i] > 0 && id[j] > 0 && ds.link(id[i], id[j]) ? 1 : 0;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a);
    id[a] = i;
  }
  ds.init(n);
  for (int r = 1; r < RLIMIT; ++r) {
    for (int s = r % 2 + 1; s < r; s += 2) {
      if (gcd(r, s) == 1) {
        int x = r * r - s * s;
        int y = 2 * r * s;
        if (x >= LIMIT || y >= LIMIT) {
          continue;
        }
        n -= gao(x, y);
        int z = r * r + s * s;
        if (z < LIMIT) {
          n -= gao(x, z);
          n -= gao(y, z);
        }
      }
    }
  }
  printf("%d\n", n);
  return 0;
}