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) { return b ? GCD(b, a % b) : a; }
long long lcm(long long a, long long b) { return a * b / GCD(a, b); }
int lb(int x) { return x & (-x); }
const long long N = 1e5 + 7;
const long long inf = 1e18 + 1;
const long long mod2 = 998244353;
const long long mod1 = 1e9 + 7;
const int P1 = 13331;
const int P2 = 131;
const double eps = 1e-6;
const double dbinf = 1e13 + 7;
const long long mod3 = 10000 + 7;
int n;
int a[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int gcd = a[1];
  for (int i = 2; i <= n; i++) {
    gcd = GCD(gcd, a[i]);
  }
  if (gcd >= 2) {
    cout << "YES\n"
         << "0\n";
    return 0;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if ((a[i] & 1) && (a[i + 1] & 1)) {
      a[i] = a[i + 1] = 0;
      ans++;
    }
  }
  for (int i = 1; i <= n - 1; i++) {
    if ((a[i] + a[i + 1]) & 1) {
      ans += 2;
      a[i + 1] = 0;
    } else if (a[i] & 1) {
      ans++;
      a[i + 1] = 0;
    }
  }
  cout << "YES\n" << ans << "\n";
  return 0;
}
