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 sys
import math

def int_arr(): return list(map(int, sys.stdin.readline().split()))

def str_arr(): return list(map(str, sys.stdin.readline().split()))

def input(): return sys.stdin.readline()

n=int(input())
arr=int_arr()
g=0
flag=1
if sum(arr)==0 or (len(arr)==1 and arr[0]<=1):
	print('no')
elif len(arr)==1 and arr[0]>=2:
	print('yes')
	print(0)
else:
	for i in range(n):
		g=math.gcd(g,arr[i])
		if g==1:
			flag=0
			break

	if flag==1:
		print('yes')
		print(0)

	else:
		j=0
		count=0
		while j<n-1:
			if arr[j]%2!=0:
				if arr[j+1]%2!=0:
					arr[j+1]=arr[j]+arr[j+1]
					count+=1
					j+=2
				else:
					count+=2
					j+=2
			else:
				j+=1
		if arr[n-1]%2!=0:
			count+=2
		print('yes')
		print(count)
