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 a[100010];
bool b[100010];
long long gcd(long long a, long long b) { return a % b ? gcd(b, a % b) : b; }
int main() {
  long long n, t;
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    b[i] = a[i] % 2;
  }
  long long ans = gcd(abs(a[1]), abs(a[2]));
  for (int i = 3; i <= n; i++) ans = gcd(ans, abs(a[i]));
  if (ans > 1)
    printf("YES\n0\n");
  else {
    int num = 0;
    for (int i = 1; i < n; i++)
      while (b[i] == 1) {
        if (b[i + 1] == 1) {
          b[i] = 0;
          b[i + 1] = 0;
        } else {
          b[i + 1] = 1;
        }
        ++num;
      }
    if (b[n] == 1) num += 2;
    printf("YES\n%d\n", num);
  }
  return 0;
}
