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) {
  long long t;
  if (a > b) {
    t = a;
    a = b;
    b = t;
  }
  while (a % b != 0) {
    t = b;
    b = a % b;
    a = t;
  }
  return b;
}
long long a[200020];
int n;
int main() {
  scanf("%d", &n);
  long long cd;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    if (i == 1)
      cd = a[1];
    else
      cd = gcd(cd, a[i]);
  }
  if (cd > 1) {
    printf("YES\n0");
    return 0;
  }
  bool flag = true;
  int count = 0;
  while (flag) {
    flag = false;
    for (int i = 1; i < n; i++) {
      if (a[i] & 1) {
        flag = true;
        count++;
        a[i] -= a[i + 1];
        a[i + 1] = a[i + 1] * 2 + a[i];
        i++;
      }
    }
  }
  if (a[n] & 1) count += 2;
  printf("YES\n%d", count);
  return 0;
}
