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.Scanner;

public class P798C
{
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++)
			arr[i] = scan.nextInt();
		int g = getGCD(arr);
		if (g != 1)
		{
			System.out.println("YES");
			System.out.println(0);
			return;
		}
		int steps = 0;
		for (int i = 0; i < n; i++)
		{
			if (i == n-1)
			{
				if (arr[i] % 2 != 0)
					steps += 2;
				continue;
			}
			if (arr[i] % 2 == 0)
				continue;
			if (oneStep(arr[i], arr[i+1]))
				steps += 1;
			else
				steps += 2;
			i++;
		}
		System.out.println("YES");
		System.out.println(steps);
	}
	private static boolean oneStep(int a, int b)
	{
		int diff = a-b;
		int sum = a+b;
		return (diff % 2 == 0 && sum % 2 == 0);
	}
	private static int getGCD(int[] arr)
	{
		int gcd = arr[0];
		for (int i = 1; i < arr.length; i++)
		{
			gcd = gcd(gcd, arr[i]);
		}
		return gcd;
	}

	private static int gcd(int a, int b)
	{
		a = Math.abs(a);
		b = Math.abs(b);
		return b == 0 ? a : gcd(b, a%b);
	}
}
