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. 
#from __future__ import division
import itertools
from fractions import gcd
from math import sqrt
from bisect import bisect_left , bisect_right
import heapq
from collections import deque , defaultdict , Counter
from itertools import combinations as C
def Ls():
	return list(raw_input())
def get(a):
	return map(a , raw_input().split())
def Int():
	return int(raw_input())
def Str():
	return raw_input()

n = input()
h = get(int)
if reduce(gcd,h) > 1: # as usual missed 1 case and no one hacked
	print 'YES\n0'
	exit(0)
ls = [i % 2 for i in h]
ans = 0;sm=0
#print ls
for i in xrange(n):
	if ls[i] == 1:
		sm += 1
	else:
		if sm % 2 == 0:
			ans += (sm/2)
		else:
			ans += (sm/2)
			ans += 2
		sm = 0

#print sm
if sm % 2 == 0:ans += (sm/2)
else:
	ans += (sm/2)
	ans += 2

print 'YES'
print ans
		




	

