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.util.*;

public class Main
{
    public static int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }
    public static void solve(int n, int a[], int temp)
    {
        int ans = 0, t;
        for (int i = 0; i < n - 1; i++)
        {
            while (a[i] % 2 != 0)
            {
                t = a[i];
                a[i] -= a[i + 1];
                a[i + 1] += t;
                ans++;
            }
        }
        if (a[n - 1] % 2 != 0)
            ans += 2;
        System.out.println("YES\n" + ans);
    }
    public static void main(String[] args)
    {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt(), temp = 0;
        int[] a = new int[100005];
        for (int i = 0; i < n; i++)
        {
            a[i] = scan.nextInt();
            temp = gcd(temp, a[i]);
        }
        if (temp > 1)
            System.out.println("YES\n" + 0);
        else
            solve(n, a, temp);
    }
}