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 MAX = 10000000 + 10;
int R[MAX], P[MAX];
bool Mark[MAX];
int pred(int x) {
  if (x == P[x]) return x;
  return P[x] = pred(P[x]);
}
void link(int a, int b) {
  if (R[a] < R[b]) {
    P[a] = b;
  } else {
    P[b] = a;
  }
  if (R[a] == R[b]) ++R[b];
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int N;
void addEdge(int u, int v) {
  if (gcd(u, v) != 1) return;
  if (Mark[u] && Mark[v]) {
    link(pred(u), pred(v));
  }
}
int main() {
  for (int i = 0; i < MAX; ++i) {
    P[i] = i;
  }
  long long hi = 0;
  scanf("%d", &N);
  for (int i = 0; i < N; ++i) {
    int x;
    scanf("%d", &x);
    if (Mark[x]) continue;
    Mark[x] = true;
    long long xx = x;
    hi = max(hi, xx);
  }
  for (long long v = 1; v * (v + 1) * 2 <= hi; ++v) {
    long long t = hi / (2 * v);
    for (long long u = v + 1; u <= t; ++u) {
      long long a = u * u - v * v;
      if (a > hi) break;
      long long b = 2 * u * v;
      long long c = u * u + v * v;
      addEdge(a, b);
      if (c <= hi) {
        addEdge(b, c);
        addEdge(a, c);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < MAX; ++i) {
    if (Mark[i] && P[i] == i) ++ans;
  }
  cout << ans << endl;
  return 0;
}