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. 

import java.io.IOException;
import java.util.Scanner;

public class Main{

    public static void main(String[] args) throws IOException {

      Scanner scanner = new Scanner(System.in);
      int n = scanner.nextInt();
      int[] a = new int[n];
      int gcd = 0;
      for(int i = 0; i < n; ++i){
          a[i] = scanner.nextInt();
          gcd = gcd(gcd, a[i]);
      }

      if(gcd > 1){
          System.out.println("YES");
          System.out.println(0);
      }else{
          int ans = 0;
          int i = 0;
          for( ; i < n - 1; ){
              if(a[i] % 2 == 1){
                  if(a[i] % 2 == 1){
                      if(a[i + 1] % 2 == 1) {
                          i += 2;
                          ++ans;
                      }else {
                          ans += 2;
                          ++i;
                      }
                  }
              }else ++i;
          }
          if(i == n - 1 && a[n - 1] % 2 == 1) ans += 2;
          System.out.println("YES");
          System.out.println(ans);
      }




    }

    private static int gcd(int a, int b){
        if(a % b == 0) return b;
        return gcd(b, a % b);
    }
}