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.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class c {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FS in = new FS(System.in);
		int n = in.nextInt();
		long[] nums = new long[n];
		for (int i = 0; i < n; i++) {
			nums[i] = in.nextInt();
		}
		long gcd = nums[0];
		for (int i = 0; i < nums.length; i++) {
			gcd = gcd(gcd, nums[i]);
		}
		if (gcd > 1) {
			System.out.println("YES");
			System.out.println(0);
			return;
		}
		int count = 0;
		PriorityQueue<Integer> spots = new PriorityQueue<Integer>();
		for (int i = 0; i < nums.length; i++) {
			if (nums[i] % 2 == 1) {
				spots.add(i);
			}
		}
		while (!spots.isEmpty()) {
			int loc = spots.poll();
			if (nums[loc] % 2 == 0)
				continue;
			if (loc < nums.length - 1 && nums[loc + 1] % 2 == 1) {
				long a = nums[loc];
				long b = nums[loc + 1];
				count++;
				nums[loc] = a - b;
				nums[loc + 1] = a + b;
			} else {
				count += 2;
				if (loc > 0) {
					long a = nums[loc - 1];
					long b = nums[loc];
					nums[loc - 1] = a - b;
					nums[loc] = a + b;
					a = nums[loc - 1];
					b = nums[loc];
					nums[loc - 1] = a - b;
					nums[loc] = a + b;
				} else {
					long a = nums[loc];
					long b = nums[loc + 1];
					nums[loc] = a - b;
					nums[loc + 1] = a + b;
					a = nums[loc];
					b = nums[loc + 1];
					nums[loc] = a - b;
					nums[loc + 1] = a + b;
				}
			}
		}
		System.out.println("YES");
		System.out.println(count);
		//System.out.println(Arrays.toString(nums));
	}

	public static long gcd(long a, long b) {
		if (a < 0 || b < 0)
			return gcd(Math.abs(a), Math.abs(b));
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	private static class FS {
		BufferedReader br;
		StringTokenizer st;

		public FS(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
