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. 
import math
from functools import reduce
'''
def euclidean_alg(a,b):
    a = abs(a)
    b = abs(b)
    if b>a:
        return euclidean_alg(b,a)
    if b == 0:
        return a
    else:
        r = a % b
        return euclidean_alg(b,r)
'''   

   
n = int(input())
s = input().split()
A = [int(i) for i in  s]
'''
if gcd(A) != 1:
    print('YES\n' + '0')
else:
    d = 0
    for i in range(len(A)-1):
        if A[i]%2 == 1 and A[i+1]%2 == 1:
            a = A[i]
            A[i] = a-A[i+1]
            A[i+1] = a + A[i+1]
            d += 1
    if d == 0 or gcd(A)==1:
        t = d
        for i in range(len(A)-1):
            if A[i]%2 == 1 or A[i+1]%2 == 1:
                a = A[i]
                A[i] = a-A[i+1]
                A[i+1] = a + A[i+1]
                a = A[i]
                A[i] = a-A[i+1]
                A[i+1] = a + A[i+1]
                d += 2
        if t == d:
            print('NO')
        else: print('YES\n' + str(d))
    else:
        print('YES\n' + str(d))
        '''
L = A.copy()
gcd = reduce(lambda x,y: math.gcd(x,y), L)        
if gcd != 1:
    print('YES\n' + '0')
else:    
    d = 0
    for i in range(len(A)-1):
        if A[i]%2 == 1 and A[i+1]%2 == 1:
            a = A[i]
            A[i] = a-A[i+1]
            A[i+1] = a + A[i+1]
            d += 1
    h = 0
    while h < len(A) and A[h] % 2 == 0:
        h += 1
    if h < len(A):
        t = d
        for i in range(len(A)-1):
            if A[i]%2 == 1 or A[i+1]%2 == 1:
                a = A[i]
                A[i] = a-A[i+1]
                A[i+1] = a + A[i+1]
                a = A[i]
                A[i] = a-A[i+1]
                A[i+1] = a + A[i+1]
                d += 2
        if t == d:
            print('NO')
        else:
            print('YES\n' + str(d))
    else:
        print('YES\n' + str(d))