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.PrintWriter;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = scan.nextInt();
		int[] a = new int[n];
		
		int g = 0;
		for(int i=0; i<n; ++i) {
			a[i] = scan.nextInt();
			g = gcd(g, a[i]);
		}
		
		if(g > 1) {
			out.println("YES");
			out.println(0);
		}
		else {
			int cnt = 0;
			for(int i=0; i<n; ++i) {
				if((a[i] & 1) == 1) {
					if(i+1 < n) {
						if((a[i+1] & 1) == 1) {
							a[i] ^= 1;
							a[i+1] ^= 1;
							cnt += 1;
						}
						else {
							a[i] ^= 1;
							cnt += 2;
						}
					}
					else {
						cnt += 2;
					}
				}
			}
			out.println("YES");
			out.println(cnt);
		}
		
		out.close();
		scan.close();
	}
	
	static int gcd(int a, int b) {
		return b == 0 ? a : gcd(b, a % b);
	}
}
