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:
		if a < b:
			a, b = b, a
		a, b = b, a % b
	return a


def main():
	n = int(input())
	a = list(map(int, input().split()))

	b = a[:]
	a.sort()

	for i in range(n - 1):
		if gcd(a[i], a[i + 1]) == 1:
			break
	else:
		print("YES\n0")
		return
	
	result = 0
	a = [0 if item % 2 == 0 else 1 for item in b]
	for i in range(n - 1):
		if a[i] == 1 and a[i + 1] == 1:
			a[i], a[i + 1] = 0, 0
			result += 1

	for i in range(n - 1):
		if a[i] == 1:
			result += 2

	if a[-1] == 1:
		result += 2

	print("YES\n%d" % result)


if __name__ == "__main__":
	main()
	# print(gcd(100, 2))

