Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given n numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given n numbers finds one that is different in evenness.

Input

The first line contains integer n (3 ≤ n ≤ 100) — amount of numbers in the task. The second line contains n space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.

Output

Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.

Examples

Input

5
2 4 7 8 10


Output

3


Input

4
1 2 1 1


Output

2
import sys
import math
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
from functools import reduce
sys.setrecursionlimit(10**6)
def inputt():
    return sys.stdin.readline().strip()
def printt(n):
    sys.stdout.write(str(n)+'\n')
def listt():
    return [int(i) for i in inputt().split()]
 
def gcd(a,b): 
   return math.gcd(a,b) 
 
def lcm(a,b): 
    return (a*b) // gcd(a,b) 

def factors(n):
    step = 2 if n%2 else 1
    return set(reduce(list.__add__,([i, n//i] for i in range(1, int(math.sqrt(n))+1, step) if n % i == 0)))

def comb(n,k):
    factn=math.factorial(n)
    factk=math.factorial(k)
    fact=math.factorial(n-k)
    ans=factn//(factk*fact)
    return ans

def is_prime(n): 
    if n <= 1: 
        return False
    if n == 2: 
        return True
    if n > 2 and n % 2 == 0: 
        return False

    max_div = math.floor(math.sqrt(n)) 
    for i in range(3, 1 + max_div, 2): 
        if n % i == 0: 
            return False
    return True
def maxpower(n,x):
    B_max = int(math.log(n, x)) + 1#tells upto what power of x n is less than it like 1024->5^4
    return B_max

#t=int(inputt())
t=1
for _ in range(t):
    n=int(inputt())
    l=listt()
    ce=[]
    co=[]
    for i in l:
        if i%2==0:
            ce.append(i)
        else:
            co.append(i)
    if (len(ce)>1 and len(co)>1) or (len(ce)==0 or len(co)==0):     
        printt("NO")  
    else:
        if len(ce)==1:
            printt(l.index(ce[0])+1)
        else:
            printt(l.index(co[0])+1)    


