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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class C {


	public static void main(String[] args) throws Exception{

		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();
		int[] a = new int[n];
		int gcd = a[0] = sc.nextInt();
		for(int i = 1; i < n; i++)
		{
			a[i] = sc.nextInt();
			gcd = GCD(gcd, a[i]);
		}
		if(gcd > 1){out.println("YES\n0");}
		else
		{
			boolean[] even = new boolean[n];
			int ev = 0;
			for(int i = 0; i < n; i++)
				if(a[i] % 2 == 0)
				{
					ev++;
					even[i] = true;
				}
			int ans = 0;
			for(int i = 0; i < n-1; i++)
				if(!even[i] && !even[i+1])
				{
					ans++;
					even[i] = even[i+1] = true;
					ev += 2;
				}
			int rem = n - ev;
			ans += 2 * rem;
			out.println("YES\n" + ans);
		}
		out.flush();
		out.close();
	}

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

	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream System){br = new BufferedReader(new InputStreamReader(System));}
		public String next() throws IOException 
		{
			while (st == null || !st.hasMoreTokens()) 
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		public String nextLine()throws IOException{return br.readLine();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public double nextDouble() throws IOException {return Double.parseDouble(next());}
		public char nextChar()throws IOException{return next().charAt(0);}
		public Long nextLong()throws IOException{return Long.parseLong(next());}
		public boolean ready() throws IOException{return br.ready();}
		public void waitForInput(){for(long i = 0; i < 3e9; i++);}
	}
}