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>
long long mod = 1000000007;
using namespace std;
long long int fast_exp(long long int base, long long int ex) {
  long long int res = 1;
  while (ex > 0) {
    if (ex % 2 == 1) res = (res * base) % mod;
    base = (base * base) % mod;
    ex /= 2;
  }
  return res % mod;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  long long int re = gcd(b, a % b);
  return re;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long ar[n];
  for (int i = 0; i < n; i++) cin >> ar[i];
  long long gc = ar[0];
  for (int i = 1; i < n; i++) gc = gcd(gc, ar[i]);
  if (gc > 1) {
    cout << "YES\n0\n";
    return 0;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (ar[i] % 2 == 0) continue;
    if (i < n - 1 && ar[i + 1] % 2) {
      ar[i] = 2;
      ar[i + 1] = 2;
      ans++;
      continue;
    } else {
      ans += 2;
      ar[i] = 2;
      ar[i + 1] = 2;
    }
  }
  cout << "YES\n";
  cout << ans << "\n";
  return 0;
}
