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 = 1000005;
int par[maxn], data[maxn], ma, vi[maxn * 10];
bool vis(int x) { return (x <= ma && vi[x]); }
int findset(int x) {
  if (par[x] != x) par[x] = findset(par[x]);
  return par[x];
}
void unionset(int x, int y) {
  int fx = findset(x);
  int fy = findset(y);
  par[fx] = fy;
}
int gcd(int x, int y) {
  if (!x) return y;
  return gcd(y % x, x);
}
inline bool cop(int x, int y) { return gcd(x, y) == 1; }
int main() {
  int n;
  scanf("%d", &n);
  ma = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", data + i);
    par[i] = i;
    ma = max(ma, data[i]);
    vi[data[i]] = i;
  }
  int upi = ceil(sqrt(1.5 * ma)), upj = ceil(sqrt(0.5 * ma));
  for (int i = 2; i <= upi; i++) {
    for (int j = (i & 1) ? 2 : 1; j < i && j <= upj; j += 2) {
      int a = i * i - j * j, b = 2 * i * j, c = i * i + j * j;
      if (cop(a, b) && vis(a) && vis(b)) {
        unionset(vi[a], vi[b]);
      }
      if (cop(a, c) && vis(a) && vis(c)) {
        unionset(vi[a], vi[c]);
      }
      if (cop(b, c) && vis(b) && vis(c)) {
        unionset(vi[b], vi[c]);
      }
    }
  }
  int res = 0;
  for (int i = 1; i <= n; i++)
    if (par[i] == i) res++;
  printf("%d\n", res);
}