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.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;

public class Main {

	static Scanner in = new Scanner(System.in);

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

		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.valueOf(st.nextToken());

		int a[] = new int[n + 1];
		st = new StringTokenizer(br.readLine());
		int now = 0;
		for (int i = 1; i <= n; i++) {
			a[i] = Integer.valueOf(st.nextToken());
			now = gcd(Math.max(now, a[i]), Math.min(now, a[i]));
		}
		System.out.println("YES");
		if (now > 1) {
			System.out.println(0);
			return;
		}
		int cnt = 0;
		for (int i = 1; i <= n - 2; i++) {
			if (a[i] % 2 == 0)
				continue;
			while (a[i] % 2 != 0) {
				int t1 = a[i];
				int t2 = a[i + 1];
				a[i] = t1 - t2;
				a[i + 1] = t1 + t2;
				cnt++;
			}
		}
		//System.out.println("cnt==" + cnt);
		while (!(a[n - 1] % 2 == 0 && a[n] % 2 == 0)) {
			int t1 = a[n - 1];
			int t2 = a[n];
			a[n - 1] = t1 - t2;
			a[n] = t1 + t2;
			//System.out.println(a[n - 1] + " " + a[n]);
			cnt++;
		}
		System.out.println(cnt);
	}

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