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.InputStreamReader;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.StringTokenizer;
import java.util.Arrays;
/***
9
57 30 28 81 88 32 3 42 25
3
6 2 4

*/
public class C{

	public static void main(String[]arg) throws IOException{
		new C().read();
	}

	public void read()throws IOException{
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;
		long[]a;
		int n,i;
		n = Integer.parseInt(in.readLine());
		st = new StringTokenizer(in.readLine());
		a = new long[n];
		for(i = 0; i < n; i++){
			a[i] = Long.parseLong(st.nextToken());
		}
		if(gcd(a) > 1) System.out.println("YES\n0");	
		else{
			System.out.println("YES\n"+countMoves(a));
		}
	}

	int countMoves(long[]a){
		int cnt = 0,i,u,v;
		for(i = 0; i < a.length-1; i++){
			if(a[i]%2 != 0 && a[i+1]%2 == 0) cnt += 2;
			else if(a[i]%2 != 0 && a[i+1]%2 != 0){
				a[i+1] = 0;
				cnt++;
			}
		}
		if(a[a.length-1]%2!=0) cnt += 2;
		return cnt;
	}

	long gcd(long[]a){
		long ans = a[0];
		int i;
		for(i = 1; i < a.length; i++){
			ans = gcd(ans,a[i]);
		}
		return ans;
	}

	long gcd(long a, long b){
		return b == 0 ? a : gcd(b, a%b);
	}

	long abs(long a){
		return a > -1 ? a: (-1)*a;
	}
}