Mike has a sequence A = [a1, a2, ..., an] of length n. He considers the sequence B = [b1, b2, ..., bn] beautiful if the gcd of all its elements is bigger than 1, i.e. <image>. 

Mike wants to change his sequence in order to make it beautiful. In one move he can choose an index i (1 ≤ i < n), delete numbers ai, ai + 1 and put numbers ai - ai + 1, ai + ai + 1 in their place instead, in this order. He wants perform as few operations as possible. Find the minimal number of operations to make sequence A beautiful if it's possible, or tell him that it is impossible to do so.

<image> is the biggest non-negative number d such that d divides bi for every i (1 ≤ i ≤ n).

Input

The first line contains a single integer n (2 ≤ n ≤ 100 000) — length of sequence A.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of sequence A.

Output

Output on the first line "YES" (without quotes) if it is possible to make sequence A beautiful by performing operations described above, and "NO" (without quotes) otherwise.

If the answer was "YES", output the minimal number of moves needed to make sequence A beautiful.

Examples

Input

2
1 1


Output

YES
1


Input

3
6 2 4


Output

YES
0


Input

2
1 3


Output

YES
1

Note

In the first example you can simply make one move to obtain sequence [0, 2] with <image>.

In the second example the gcd of the sequence is already greater than 1. 
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  int n;
  cin >> n;
  long long a[n], g;
  for (int i = 0; i < n; i++) cin >> a[i];
  g = a[0];
  for (int i = 1; i < n; i++) g = gcd(g, a[i]);
  if (g > 1) {
    cout << "YES\n";
    cout << "0\n";
    return 0;
  }
  int step = 0;
  for (int i = 0; i < n; i++) {
    if (i == n - 1 && a[i] % 2) {
      step += 2;
      break;
    } else {
      if (a[i] % 2) {
        if (a[i + 1] % 2)
          step++;
        else
          step += 2;
        a[i] = 0;
        a[i + 1] = 0;
      }
    }
  }
  cout << "YES\n";
  cout << step << endl;
  return 0;
}
