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>
const int INF = 2000000000;
const double EPS = 1e-6;
using namespace std;
int v[100010], n, cnt;
bool isEven[100010];
int gcd(int a, int b) {
  a %= b;
  if (a == 0) return b;
  return gcd(b, a);
}
int allGcd() {
  int base = gcd(v[0], v[1]);
  for (int i = (int)(2); i < (int)(n); ++i) base = gcd(base, v[i]);
  return base;
}
int main() {
  scanf("%d", &n);
  for (int i = (int)(0); i < (int)(n); ++i)
    scanf("%d", &v[i]), isEven[i] = (v[i] & 1 ? false : true);
  if (allGcd() == 1) {
    for (int i = (int)(0); i < (int)(n - 1); ++i) {
      if (!isEven[i] && !isEven[i + 1]) {
        isEven[i] = isEven[i + 1] = true;
        cnt++;
      }
    }
    for (int i = (int)(0); i < (int)(n - 1); ++i) {
      if (isEven[i] ^ isEven[i + 1]) {
        isEven[i] = isEven[i + 1] = true;
        cnt += 2;
      }
    }
  }
  puts("YES");
  cout << cnt << endl;
  return 0;
}
