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 MikeAndgcdProblems {

	public static void main(String args[]) {
		int gcd = 0;
		int ans = 0;
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int a[] = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = scanner.nextInt();
		}
		gcd = gcd(a[0], a[1]);
		for (int j = 2; j < n; j++) {
			gcd = gcd(gcd, a[j]);
		}
		System.out.println("YES");
		if (gcd > 1) {
			System.out.println("0");
			return;
		}
		for (int k = 0; k < n - 1; k++) {
			while (a[k] % 2 != 0) {
				ans++;
				int ta = a[k] - a[k + 1];
				int tb = a[k] + a[k + 1];
				a[k] = ta;
				a[k + 1] = tb;
			}
		}
		while (a[n - 1] % 2 != 0) {
			ans++;
			int ta = a[n - 2] - a[n - 1];
			int tb = a[n - 2] + a[n - 1];
			a[n - 1] = ta;
			a[n - 2] = tb;
		}
		System.out.println(ans);
	}

	// 最大公约数
	public static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		} else {
			return gcd(b, a % b);
		}
	}
}
