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 maxn = 100005;
const int inf = 0x3f3f3f3f;
int gcd(int a, int b) {
  int c;
  while (b) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
int a[maxn], n;
int main() {
  scanf("%d", &n);
  int g = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    g = gcd(a[i], g);
  }
  printf("YES\n");
  if (g > 1) {
    printf("0\n");
    return 0;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if ((a[i] & 1) == 0)
      continue;
    else {
      if (i + 1 < n) {
        if ((a[i + 1] & 1) == 1) {
          ans += 1;
          i++;
        } else {
          ans += 2;
          i++;
        }
      } else
        ans += 2;
    }
  }
  printf("%d\n", ans);
  return 0;
}
