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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main
{

    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void main(String[] args)
    {
        setup();
        xuly();
    }

    private static int n;
    private static int[] a;

    private static void setup()
    {
        FastReader fastReader = new FastReader();
        n = fastReader.nextInt();
        a = new int[n];
        for (int i = 0; i < n; i ++)
            a[i] = fastReader.nextInt();
    }

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

    private static void xuly()
    {
        System.out.print("YES\n");
        int gcdA = Math.abs(a[0]);
        for (int i = 1; i < n; i ++)
            gcdA = gcd(gcdA, Math.abs(a[i]));
        if (gcdA != 1)
            System.out.print(0);
        else
        {
            int cnt = 0;
            for (int i = 0; i < n; i ++)
                if ((a[i] & 1) != 0)
                {
                    if (i == n - 1)
                        cnt += 2;
                    else if ((a[i + 1] & 1) != 0) {
                        cnt ++;
                        a[i + 1] ++;
                    }
                    else
                        cnt += 2;
                }
            System.out.print(cnt);
        }
    }
}
