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.PrintStream;
import java.util.Scanner;

public class C {
	
	private static final PrintStream out = System.out;
	private static final Scanner in = new Scanner(System.in);
	
	private static int gcd(int a, int b) {
		int t;
		while (b > 0) {
			t = b;
			b = a%b;
			a = t;
		}
		return a;
	}
	
	public static void main(String[] args) {
		int n = in.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; ++i)
			a[i] = in.nextInt();
		int g = a[0];
		for (int i = 1; i < n; ++i)
			g = gcd(g, a[i]);
		int ans = 0;
		if (g == 1) {
			for (int i = 0; i < n; ++i)
				a[i] = a[i]%2;
			for (int i = 0; i < n; ++i) {
				if (a[i] == 1) {
					if (i == n - 1) {
						ans += 2;
					} else if (a[i + 1] == 0) {
						ans += 2;
						++i;
					} else if (a[i + 1] == 1) {
						++ans;
						++i;
					}
				}
			}
		}
		out.println("YES");
		out.println(ans);
	}
	
}
