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. 
/**
 * Created by mypc on 21/4/17.
 */

import java.io.*;
import java.util.*;

public class Q3 {
    static PrintWriter pw = new PrintWriter(System.out);

    public static void main(String args[]) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        //BufferedReader br = new BufferedReader(new FileReader("testcase.txt"));
        StringTokenizer st;
        //st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(br.readLine());
        int a[] = new int[n];
        st = new StringTokenizer(br.readLine());
        for(int i = 0 ; i < n ; i++)
            a[i] = Integer.parseInt(st.nextToken());

        int g = gcd(a);
        if(g != 1)
            pw.println("YES\n0");
        else{
            int answer = 0;
            for(int i = 0 ; i < n ; i++)
            {
                if(a[i] % 2 == 1)
                {
                    if(i+1 < n && a[i+1]%2 == 1) {
                        answer++;
                        a[i]++;
                        a[i+1]++;
                    }
                    else
                        answer += 2;
                }
            }
            pw.println("YES\n"+answer);
        }
        pw.flush();
    }
    public static int gcd(int a[])
    {
        int g = a[0];
        for(int i = 1 ; i < a.length ; i++)
        {
            g = gcd(g,a[i]);
            if(g == 1)
                return 1;
        }
        return g;
    }
    public static int gcd(int a,int b)
    {
        if(a < b)
        {
            int temp = a;
            a = b;
            b = temp;
        }
        if(b == 0)
            return a;
        else return gcd(b,a%b);
    }
}
