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;
using ll = long long;
const int mod = 1e9 + 7;
const int inf = (1 << 30) - 1;
const ll infll = (1LL << 61) - 1;
int N, g;
template <typename T>
T gcd(T a, T b) {
  return (b ? gcd(b, a % b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return a / (gcd(a, b)) * b;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> N;
  vector<int> a(N);
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    g = gcd(a[i], g);
  }
  if (g > 1) {
    cout << "YES\n0\n";
  } else {
    int res = 0;
    for (int i = 0; i + 1 < N; i++) {
      if (a[i] % 2 == 1) {
        if (a[i + 1] % 2 == 0)
          res += 2, a[i]++;
        else
          res++, a[i]++, a[i + 1]++;
      }
    }
    if (a[N - 1] % 2 == 1) res += 2;
    cout << "YES\n" << res << "\n";
  }
}
