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;
const int N = 1e5 + 5;
int gcd(int n, int m) {
  int t = 1;
  while (t) {
    t = n % m;
    n = m;
    m = t;
  }
  return n;
}
int a[N];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int cnt = 0;
  int t;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i != 0)
      t = gcd(t, a[i]);
    else
      t = a[i];
  }
  if (t > 1) {
    cout << "YES" << endl;
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    if (a[i] & 1 && a[i + 1] & 1) cnt++, a[i] = a[i + 1] = 2;
  }
  for (int i = 0; i < n - 1; i++) {
    if ((a[i] & 1 && a[i + 1] % 2 == 0) || (a[i] % 2 == 0 && a[i + 1] & 1))
      cnt += 2, a[i] = a[i + 1] = 2;
  }
  cout << "YES" << endl;
  cout << cnt << endl;
  return 0;
}
