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.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.Map;
import java.io.PrintWriter;
import java.util.Map.Entry;
import java.util.regex.*;
/*created by aditya*/
public class adi {
        private boolean[] marked;
        public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        //PrintWriter me = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] vals=br.readLine().split(" ");
        int n=Integer.parseInt(vals[0]);
        vals=br.readLine().split(" ");
        int[] arr=new int[n];
        for(int i=0;i<n;i++){
          arr[i]=Integer.parseInt(vals[i]);
        }
        int g=arr[0];
        for(int i=1;i<n;i++){
          g=gcd(arr[i],g);
        }
        if(g>1){
          System.out.println("YES");
          System.out.println("0");
          return;
        }
        int ans=0;
        for(int i=0;i<n;i++){
          if(arr[i]%2==0)
            continue;
          if(i+1<n&&arr[i+1]%2==1){
            ans++;
            i++;
          }
          else
            ans+=2;
        }
        System.out.println("YES");
        System.out.println(ans);

  }

public static void dfs(ArrayList<ArrayList<Integer>> adjLists,boolean[] visited,int v){
  visited[v]=true;
  //System.out.print(v+" ");
  for(int w:adjLists.get(v)){
    if(!visited[w]){
      dfs(adjLists,visited,w);
    }
  }
}
public static int gcd(int p, int q) {
          if (q == 0)
            return p;
          return gcd(q, p % q);
        }

public static boolean isprime(long number){
            for(int i=2; i<=number/2; i++){
                if(number % i == 0){
                    return false;
                }
            }
            return true;
      }

public static ArrayList<Integer> sieve(int n){
  boolean[] isPrime = new boolean[n+1];
  ArrayList<Integer> al=new ArrayList<Integer>();
        for (int i = 2; i <= n; i++)
            isPrime[i] = true;
        for (int factor = 2; factor*factor <= n; factor++) {
            if (isPrime[factor]) {
                for (int j = factor; factor*j <= n; j++)
                    isPrime[factor*j] = false;
            }
        }
        int primes = 0;
        for (int i = 2; i <= n; i++) {
            if (isPrime[i])
              al.add(i);
            }
        return al;
        }
   }
