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. 
#!/usr/bin/env python3
from sys import stdin,stdout
from math import *

def ri():
    return map(int, stdin.readline().split())

def GCD(a):
    ans = 0
    for e in a:
        ans = gcd(ans, e)
    return ans

def even(i):
    return (i+1)%2

def odd(i):
    return (i)%2

n = int(input())

a = list(ri())
print("YES")
if GCD(a) > 1:
    print(0)
    exit()

count = 0
for i in range(n-2):
    if even(a[i]):
        if even(a[i+1]):
            continue
        else:
            if odd(a[i+2]):
                continue
            else:
                count += 2
                a[i+1] +=1
    else:
        if even(a[i+1]):
            count += 2
            continue
        else:
            count += 1
            a[i+1] += 1

if even(a[n-2]) and even(a[n-1]):
    pass
if even(a[n-2]) and odd(a[n-1]):
    count += 2
    pass
if even(a[n-1]) and odd(a[n-2]):
    count += 2
    pass
if odd(a[n-2]) and odd(a[n-1]):
    count += 1
    pass

print(count)

