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.*;
public class C798 
{
    static int find(int a,int b)
    {
        if(b==0)
            return a;
        else
            return find(b,a%b);
    }
    public static void main(String args[])throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        String s[]=br.readLine().split(" ");
        int arr[]=new int[n];int fl=0;int gcd=0;
        for(int i=0;i<n;i++)
        {
            arr[i]=Integer.parseInt(s[i]);
            if(i==0)
                gcd=arr[0];
            else
                gcd=find(gcd,arr[i]);
        }
        //System.out.println(gcd);
        if(gcd!=1)
        {
            System.out.println("YES");
            System.out.println("0");
        }
        else
        {
            int par[]=new int[n];
            for(int i=0;i<n;i++)
            {
                if(arr[i]%2!=0)
                    par[i]=1;
            }
            int i=n-1,tot=0;
            while(i>=0)
            {
                if(i!=0)
                {
                    if(par[i]==1)
                    {
                        if(par[i-1]==1)
                        {
                            tot++;
                            par[i-1]=0;
                            par[i]=0;
                            i-=2;
                        }
                        else
                        {
                            tot+=2;
                            par[i]=0;
                            i-=2;
                        }
                    }
                    else
                        i--;
                }
                else
                {
                    if(par[0]==1)
                    {
                        tot+=2;
                        par[0]=0;
                        i--;
                    }
                    else
                        i--;
                }
            }
            System.out.println("YES");
            System.out.println(tot);
        }
    }

}


