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. 
//package codeforces;

import java.util.*;

public class P798C {
   
   public static boolean isOdd(int x){
	   if (x%2 != 0) return true;
	   return false;
   }
   
   public static int gcd(int a ,int b){
	   if (b == 0) return a;
	   else return gcd(b, a % b);
   }
   
   public static void main(String[] args){
	   Scanner input = new Scanner(System.in);
	   int n = input.nextInt();
	   
	   int[] a = new int[n+10];
	   for (int i = 0; i < n ; i++ ){
		   a[i] = input.nextInt();
	   }
	   
	   int counter = 0;
	   int gcd = a[0];
	   for (int i = 1 ; i <n ; i++) gcd = gcd(gcd , a[i]);
	   
	   
	   if (gcd <= 1) { 
	   
	  
	     for (int i = 0 ; i < n-1 ; i++ ){
		    if (isOdd(a[i]) == true && isOdd(a[i+1]) == true) {
		    	++counter;
		    	int x = a[i] - a[i+1];
		    	int y = a[i] + a[i+1];
		    	a[i] = x;
		    	a[i+1] = y;
		    };
	     }
	     
	     for (int i = 0 ; i < n-1; i++) {
	    	 if (isOdd(a[i]) == true && isOdd(a[i+1]) == false) {
	    		 counter+=2;
	    	 }
	     }
	     
	     if (isOdd(a[n-1]) == true && isOdd(a[n-2]) == false) counter+=2;
	   }
	   
	   System.out.println("YES");
	   System.out.println(counter);
	   
	   input.close();
	   
   }
}
