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;

public class Other {
	
	static int n;
	
	static int[] arr;
	
	public static void main(String[] args) throws NumberFormatException, IOException {
		
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		
		n = Integer.parseInt(in.readLine());
		
		arr = new int[n];
		
		String[] line = in.readLine().split("\\s+");
		
		int g = 0;
		
		for(int i=0; i<n; i++) {
			arr[i] = Integer.parseInt(line[i]);
			
			g = gcd(g, arr[i]);
		}
		
		System.out.println("YES");
		
		if(g > 1) {
			System.out.println(0);
			return;
		}
		
		
		int rel = 0;
		int ans = 0;
		
		for(int i=0; i<n; ) {
			
			int j = i;
			
			if(arr[j] % 2 == 1) {
				
				rel = 0;
								
				while(j < n && arr[j] % 2 == 1) {
					rel++;
					j++;
				}
				
				ans += rel/2;
				
				if(rel % 2 == 0) {
					
				}
				else if(j +1 == n && i-1==0) {
					
				}
				else {
					ans += 2;
				}
				
				
		
			}
			
			i = ++j;
			
			
				
		}
		

		
		System.out.println(ans);
		
		

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

