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 re
from math import gcd
'''
def gcd(a,b):
    if a<b:
        t=a
        a=b
        b=t
    r=a%b
    while r>0:
        a=b
        b=r
        r=a%b
    return b
    '''
class Solution(object):
    def sol(self):
        n=int(input())
        nums = [int(i) for i in input().split()]
        s=''
        g=0
        for i in range(n):
            g=gcd(g,nums[i])
            s+='0' if nums[i]%2==0 else '1'
                
        def gcdi(id):
            if id==n-1:
                return nums[n-1]
            else:
                return gcd(gcdi(id+1),nums[id])
        ones=re.split('0+',s)
        res=0
        for s1 in ones:
            res+=len(s1)//2 if len(s1)%2==0 else len(s1)//2+2
        print('YES')
        if g>1 :
            print('0')
        else :
            print(res)
so=Solution()
so.sol()