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>
#pragma comment(linker, "/STACK:10240000000,10240000000")
using namespace std;
int a[100005];
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  int n;
  while (cin >> n) {
    int gc = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      gc = gcd(max(gc, a[i]), min(gc, a[i]));
      if (a[i] & 1)
        a[i] = 1;
      else
        a[i] = 0;
    }
    puts("YES");
    if (gc > 1)
      puts("0");
    else {
      int res = 0;
      int cnt = 0;
      for (int i = 0; i < n; i++) {
        if (a[i] == 1)
          cnt++;
        else {
          res += (cnt) / 2 + 2 * (cnt & 1);
          cnt = 0;
        }
      }
      if (cnt != 0) res += (cnt) / 2 + 2 * (cnt & 1);
      cout << res << endl;
    }
  }
  return 0;
}
