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 gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 1e5 + 10;
int arr[MAXN];
int main() {
  int n;
  while (cin >> n) {
    int t;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
      if (i == 0)
        t = arr[i];
      else
        t = gcd(t, arr[i]);
    }
    if (t > 1) {
      cout << "YES" << endl << 0 << endl;
      continue;
    }
    int ans = 0;
    for (int i = 0; i < n - 1; ++i) {
      while (1) {
        if (arr[i] % 2 == 0) break;
        ans++;
        int a1 = arr[i] - arr[i + 1], a2 = arr[i] + arr[i + 1];
        arr[i] = a1;
        arr[i + 1] = a2;
      }
    }
    if (arr[n - 1] % 2 == 1) ans += 2;
    cout << "YES" << endl << ans << endl;
  }
  return 0;
}
