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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;

import sun.rmi.transport.proxy.CGIHandler;


public class A {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n = nextInt();
		long[]a = new long[n+1];
		for (int i = 1; i <= n; i++) {
			a[i] = nextInt();
		}
		long gcd = 0;
		for (int i = 1; i <= n; i++) {
			gcd = gcd(gcd, a[i]);
		}
		if (gcd > 1) {
			System.out.println("YES");
			System.out.println(0);
			return;
		}
		int ans = 0;
		for (int i = 1; i < n; i++) {
			if (a[i] % 2 != 0) {
				int cnt = 1;
				if (a[i+1] % 2 == 0)
					cnt = 2;
				ans += cnt;
				while (cnt-->0) {
					long ai = a[i]-a[i+1];
					long ai1 = a[i]+a[i+1];
					a[i] = ai;
					a[i+1] = ai1;
				}
			}
		}
		if (a[n] % 2 != 0)
			ans += 2;
		System.out.println("YES");
		System.out.println(ans);
		pw.close();
	}
	private static long gcd(long a, long b) {
		if (b==0)
			return a;
		return gcd(b, a % b);
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}