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 bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline

from itertools import accumulate
from functools import lru_cache

M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
 
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]


# sys.setrecursionlimit(10 ** 6)


n = val()
l = li()






# @lru_cache(None)
# def dp(i = 1, odd = l[0] & 1):
#     if i == n:
#         return 2 if odd else 0
    
#     if odd:
#         if l[i] & 1:
#             return 1 + dp(i + 1, 0)
#         else:
#             return 2 + dp(i + 1, 0)
#     else:
#         if l[i] & 1:
#             return min(dp(i + 1, 1), 2 + dp(i + 1, 0))
#         else:return dp(i + 1, 0)

gc = 0
for i in l:gc = math.gcd(gc, i)
print('YES')
if gc > 1:
    print(0)
    exit()


dp = [[0] * 2 for i in range(n + 1)]

dp[n][0] = 0
dp[n][1] = 2
for i in range(n - 1, -1, -1):
    dp[i][0] = min(dp[i + 1][1], 2 + dp[i + 1][0]) if l[i] & 1 else dp[i + 1][0]
    dp[i][1] = dp[i + 1][0] + (1 if l[i] & 1 else 2)

print(dp[1][l[0]&1])

