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.*;

public class C1
{
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		boolean[] bits = new boolean[n];
		int gcd = 0;
		for(int i = 0; i < n; i++)
		{
			int num = Integer.parseInt(st.nextToken());
			gcd = gcd(gcd, num);
			bits[i] = num%2 == 1;
		}
		int moves = 0, i = 0;
		while(i < n)
		{
			if(bits[i])
			{
				int j = i;
				while(j < n-1 && bits[j+1])
					j++;
				int len = j-i+1;
				moves += (int)(Math.ceil(len/2.0)) + (len%2 == 1 ? 1 : 0);
				i = j;
			}
			i++;
		}
		pw.println("YES");
		pw.println(gcd != 1 ? 0 : moves);
		br.close();
		pw.close();
		System.exit(0);
	}
	public static int gcd(int a, int b)
	{
		return b == 0 ? a : gcd(b, a%b);
	}
}