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 fractions import gcd
n = input()
a = map(int, raw_input().split())
b = sorted(a)
g = reduce(gcd, b)
if g > 1:
    print "YES"
    print 0
else:
    count = 0
    eo = [(0 if i % 2  == 0 else 1) for i in a]
    for i in range(n):
        if eo[i] == 1:
            if i < n - 1 and eo[i + 1] == 1:
                count += 1
                eo[i], eo[i + 1] = 0, 0
                i += 1
            elif i < n - 1 and eo[i + 1] == 0:
                count += 2
                eo[i], eo[i + 1] = 0, 0
                i += 1
            elif eo[i - 1] == 1:
                count += 1
                eo[i], eo[i - 1] = 0, 0
                i += 1
            elif eo[i - 1] == 0:
                count += 2
                eo[i], eo[i - 1] = 0, 0
                i += 1
    print "YES"
    print count