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 maxx = int(sqrt(10000000 * 3 / 2.0));
const int maxy = int(sqrt(10000000 / 2.0));
int n, a[10000000 + 5], father[1000005], ans;
void init() {
  scanf("%d", &n);
  int i, x;
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    a[x] = i;
  }
}
int gcd(int x, int y) {
  int t;
  while (y) {
    t = y;
    y = x % y;
    x = t;
  }
  return x;
}
int getfather(int x) {
  int t, temp, tmp;
  tmp = x;
  while (father[tmp] != tmp) tmp = father[tmp];
  temp = tmp;
  tmp = x;
  while (father[tmp] != temp) {
    t = father[tmp];
    father[tmp] = temp;
    tmp = t;
  }
  return temp;
}
void merge(int x, int y) {
  int fx, fy;
  fx = getfather(x);
  fy = getfather(y);
  if (fx != fy) {
    father[fy] = fx;
    ans--;
  }
}
void work() {
  int i, x, y;
  int aa, bb, cc;
  for (i = 1; i <= n; i++) father[i] = i;
  ans = n;
  for (x = 1; x <= maxx; x++)
    for (y = (x & 1) + 1; y < x && y <= maxy; y += 2)
      if (gcd(x, y) == 1) {
        aa = x * x - y * y;
        bb = 2 * x * y;
        cc = x * x + y * y;
        if (aa > 10000000 || bb > 10000000) continue;
        if (a[aa] && a[bb]) merge(a[aa], a[bb]);
        if (cc <= 10000000 && a[cc]) {
          if (a[aa]) merge(a[aa], a[cc]);
          if (a[bb]) merge(a[bb], a[cc]);
        }
      }
  printf("%d\n", ans);
}
int main() {
  init();
  work();
  return 0;
}