Given is a sequence of N digits a_1a_2\ldots a_N, where each element is 1, 2, or 3. Let x_{i,j} defined as follows:

* x_{1,j} := a_j \quad (1 \leq j \leq N)
* x_{i,j} := | x_{i-1,j} - x_{i-1,j+1} | \quad (2 \leq i \leq N and 1 \leq j \leq N+1-i)



Find x_{N,1}.

Constraints

* 2 \leq N \leq 10^6
* a_i = 1,2,3 (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:


N
a_1a_2\ldotsa_N


Output

Print x_{N,1}.

Examples

Input

4
1231


Output

1


Input

10
2311312312


Output

0
import sys
import math
import heapq
mod=10**9+7
inf=float("inf")
from math import sqrt, ceil
from collections import deque, Counter, defaultdict, OrderedDict
#すべてのkeyが用意されてる defaultdict(int)で初期化
#順序を保ったdict
input=lambda: sys.stdin.readline().strip()
sys.setrecursionlimit(11451419)
from decimal import ROUND_HALF_UP,Decimal  #変換後の末尾桁を0や0.01で指定
  #Decimal((str(0.5)).quantize(Decimal('0'), rounding=ROUND_HALF_UP))
from functools import lru_cache
#メモ化再帰defの冒頭に毎回 @lru_cache(maxsize=10**10)
#引数にlistはだめ
#######ここまでテンプレ#######
#ソート、"a"+"b"、再帰ならPython3の方がいい
#######ここから天ぷら########
n=int(input())
s=input()
A=[int(i)-1 for i in s]

RUI2=[pow(2,i) for i in range(21)]
import bisect as bi
@lru_cache(maxsize=10**10)
def tan(n):
    if n==0:return [1]
    if n==1:return [1,1]
    if n==2:return [1,0,1]
    i= bi.bisect_right(RUI2,n)-1
    return tan(n-pow(2,i)) + [0]*(pow(2,i+1)-n-1) + tan(n-pow(2,i))


# for i in range(n-1):
#     if A[i]==A[i+1]: A[i]=0
#     elif (A[i]^A[i+1])&1 : A[i]=1
#     else:A[i]=2
# for i in range(n-2):
#     if (A[i]^A[i+1])&1 : A[i]=1
#     else:A[i]=0

# print(A[:-2],tan(n-3))
if A.count(1)!=0:
    A=[i%2!=1 for i in A]
    a=[i&j for i,j in zip(A,tan(n-1))]
    # print(a)
    print(1 if sum(a)%2==1 else 0)
else:
    a=[(i//2)&j for i,j in zip(A,tan(n-1))]
    #print(a,tan(4))
    print(2 if sum(a)%2==1 else 0)
    
