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 mx = 10000002;
int pos[mx];
int u[1000003];
int a[1000003];
int p[1000003];
int n;
inline int gcd(int x, int y) {
  while (x && y)
    if (x >= y)
      x %= y;
    else
      y %= x;
  return x + y;
}
int ans;
inline int parent(int x) {
  int who = p[x];
  while (p[who] != who) who = p[who];
  while (p[x] != x) {
    int tmp = p[x];
    p[x] = who;
    x = tmp;
  }
  return who;
}
inline void unite(int x, int y) {
  if (gcd(a[x], a[y]) != 1) return;
  int p1 = parent(x);
  int p2 = parent(y);
  if (p1 != p2) {
    if ((x ^ y) & 1)
      p[p1] = p2;
    else
      p[p2] = p1;
    --ans;
  }
}
int main() {
  scanf("%d", &n);
  for (int(i) = 0; (i) < (n); ++(i)) {
    int x;
    scanf("%d", &x);
    a[i + 1] = x;
    pos[x] = i + 1;
    p[i + 1] = i + 1;
  }
  ans = n;
  for (int i = 1; i < mx; ++i)
    for (int j = i + 1; j < mx; ++j) {
      long long b = (long long)2 * i * j;
      long long a = (long long)j * j - (long long)i * i;
      long long c = (long long)i * i + (long long)j * j;
      if (c > mx && b > mx) break;
      if (b < mx && pos[b]) {
        if (a < mx && pos[a]) unite(pos[a], pos[b]);
        if (c < mx && pos[c]) unite(pos[c], pos[b]);
      }
      if (a < mx && c < mx && pos[a] && pos[c]) unite(pos[a], pos[c]);
    }
  cout << ans << endl;
}