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>
int n, m, i, ogr, ans;
bool pt;
long long x, y, zz;
int num[10000000 + 10];
int e[10000000], next[10000000];
int head[1000010];
bool u[1000010];
int nod(int a, int b) {
  if (!b) return a;
  return nod(b, a % b);
}
void addreb(int x, int y) {
  if (y > n) {
    printf("fgfh");
    return;
  }
  m++;
  e[m] = y;
  next[m] = head[x];
  head[x] = m;
  m++;
  e[m] = x;
  next[m] = head[y];
  head[y] = m;
}
void dfs(int k) {
  int v = head[k];
  u[k] = 1;
  while (v != 0) {
    if (!u[e[v]]) dfs(e[v]);
    v = next[v];
  }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    num[x] = i;
  }
  for (x = 1;; x++) {
    ogr = 10000000 / (x * 2);
    if (x + 1 > ogr) break;
    pt = true;
    for (y = x + 1; y <= ogr; y += 2) {
      if (y * y - x * x > 10000000) break;
      if (num[x * y * 2]) {
        if (nod(x, y) > 1) continue;
        if (num[y * y - x * x]) addreb(num[x * y * 2], num[y * y - x * x]);
        if (pt) {
          zz = x * x + y * y;
          if (zz > 10000000)
            pt = false;
          else {
            if (num[zz]) {
              if (num[x * y * 2]) addreb(num[x * y * 2], num[zz]);
              if (num[y * y - x * x]) addreb(num[y * y - x * x], num[zz]);
            }
          }
        }
      } else if (pt) {
        zz = x * x + y * y;
        if (zz > 10000000)
          pt = false;
        else {
          if (num[zz]) {
            if (nod(x, y) > 1) continue;
            if (num[x * y * 2]) addreb(num[x * y * 2], num[zz]);
            if (num[y * y - x * x]) addreb(num[y * y - x * x], num[zz]);
          }
        }
      }
    }
  }
  ans = 0;
  for (i = 1; i <= n; i++) {
    if (!u[i]) {
      ans++;
      dfs(i);
    }
  }
  printf("%d", ans);
  return 0;
}