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;
int several[130000];
int n;
int gcd(int n, int m) {
  if (n < m) {
    n = m + n;
    m = n - m;
    n = n - m;
  }
  if (m == 0) return n;
  return gcd(m, n % m);
}
int main() {
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &several[i]);
  int a = several[1];
  int b = several[2];
  int c = gcd(a, b);
  int i;
  for (i = 3; i <= n; i++) {
    if (c <= 1) break;
    c = gcd(c, several[i]);
  }
  if (c > 1) {
    printf("YES\n");
    printf("0");
  } else {
    for (int i = 1; i <= n - 1;) {
      if (several[i] % 2 != 0) {
        if (several[i + 1] % 2 != 0) {
          several[i]++;
          several[i + 1]++;
          cnt++;
        } else {
          several[i]++;
          cnt = cnt + 2;
        }
        i = i + 2;
      } else {
        i++;
      }
    }
    if (several[n] % 2 != 0) {
      cnt = cnt + 2;
    }
    printf("YES\n");
    printf("%d", cnt);
  }
}
