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.StringTokenizer;


public class C {
	
	static class Scanner{
		BufferedReader br=null;
		StringTokenizer tk=null;
		public Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		public String next() throws IOException{
			while(tk==null || !tk.hasMoreTokens()) {
				String cad = br.readLine();
				if (cad == null)
					return null;
				tk=new StringTokenizer(cad);
			}
			return tk.nextToken();
		}
		public int nextInt() throws NumberFormatException, IOException{
			return Integer.valueOf(next());
		}
		public long nextLong() throws NumberFormatException, IOException{
			return Long.valueOf(next());
		}
		public double nextDouble() throws NumberFormatException, IOException{
			return Double.valueOf(next());
		}
	}
	
	static int gcd(int a, int b) {
		a = Math.abs(a);
		b = Math.abs(b);
		if (b == 0)
			return a;
		return gcd(b, a % b);
	}
	
	static long[] array;
	
	public static void main(String args[]) throws NumberFormatException, IOException {
		Scanner sc = new Scanner();
		int N = sc.nextInt();
		int gcd = 0;
		array = new long[N];
		for(int i = 0; i < N; i++) {
			array[i] = sc.nextInt();
			gcd = gcd((int)array[i], gcd);
		}
		System.out.println("YES");
		if (gcd > 1)
			System.out.println(0);
		else {
			int ans = 0;
			for(int i = 0; i < N - 1; i++) {
				if ((array[i] % 2 == 1) && (array[i + 1] % 2 == 1)) {
					long f = array[i] - array[i + 1];
					long s = array[i] + array[i + 1];
					array[i] = f;
					array[i + 1] = s;
					ans++;
				}
			}
			for(int i = 0; i < N; i++)
				if (array[i] % 2 == 1)
					ans += 2;
			System.out.println(ans);
		}
	}

}
