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):
    if a == 0:
        return b
    if b == 0:
        return a

    return gcd(b, a % b)


def gcd_of_list(a):
    a = a[::-1]
    partial_gcd = 0
    while a:
        partial_gcd = gcd(partial_gcd, a.pop())
    return partial_gcd


def get_odd_chunk_lengths(a):
    a = a[::-1]

    chunk_lengths_arr = []
    curr_chunk_length = 0

    while a:
        next_num = a.pop()

        if next_num % 2 == 1:
            curr_chunk_length += 1

        if next_num % 2 == 0 and curr_chunk_length != 0:
            chunk_lengths_arr.append(curr_chunk_length)
            curr_chunk_length = 0

    if curr_chunk_length != 0:
        chunk_lengths_arr.append(curr_chunk_length)

    return chunk_lengths_arr


def num_ops_to_evenify_odd_chunk(chunk_length):
    if chunk_length % 2 == 0:
        return chunk_length // 2
    else:
        return chunk_length // 2 + 2


input()
a = [int(i) for i in input().split()]

if gcd_of_list(a) > 1:
    print("YES")
    print(0)
else:
    print("YES")
    print(sum([num_ops_to_evenify_odd_chunk(chunk_length)
               for chunk_length in get_odd_chunk_lengths(a)]))
