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. 

def gcd(a, b):
    while b > 0:
        a, b = b, a % b
    return a


def gcd_list(l):
    n = len(l)
    if n == 1:
        return l[0]
    return gcd(gcd_list(l[0:n//2]), gcd_list(l[n//2:]))


n = int(input().strip())
a = [int(x) for x in input().strip().split()]

print('YES')
if gcd_list(a) > 1:
    print(0)
else:
    a.append(0)
    ops = 0
    odds = 0
    for x in a:
        if x % 2 == 1:
            odds += 1
        elif odds > 0:
            ops += (odds // 2) + (2 if odds % 2 == 1 else 0)
            odds = 0

    print(ops)

