Problem:
Score : 700 points

Problem Statement

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}.

Sample Input 1

4
1231

Sample Output 1

1

x_{1,1},x_{1,2},x_{1,3},x_{1,4} are respectively 1,2,3,1.

x_{2,1},x_{2,2},x_{2,3} are respectively |1-2| = 1,|2-3| = 1,|3-1| = 2.

x_{3,1},x_{3,2} are respectively |1-1| = 0,|1-2| = 1.

Finally, x_{4,1} = |0-1| = 1, so the answer is 1.

Sample Input 2

10
2311312312

Sample Output 2

0

Solution in Python (3.4.3) (judge verdict: Accepted)

import sys
INF = 10 ** 9
MOD = 10 ** 9 + 7
from collections import deque
sys.setrecursionlimit(100000000)

fact2 = [0] * 1000011
def cnt2(n):
    if n%2 == 1:
        return 0
    else:
        fact2[n] = 1 + fact2[n//2]
        return fact2[n]

factorial = [0,0]
for i in range(2,1000010):
    factorial.append(factorial[-1] + cnt2(i))

def nck(n,k):
    return factorial[n] - factorial[k] - factorial[n - k]

def nxor(x,n):
    if n > 0:
        return 0
    else:
        return x


def main():
    n = int(input())
    a = [int(i) - 1 for i in list(input())]
    
    cumxor = 0
    amod = [i%2 for i in a]
    for i in range(n):
        cumxor ^= nxor(amod[i],nck(n - 1,i))
    
    if cumxor == 1:
        ans = 1
    else:
        if 1 in a:
            ans = 0
        else:
            cumxor = 0
            adiv = [i//2 for i in a]
            for i in range(n):
                cumxor ^= nxor(adiv[i],nck(n - 1,i))
            if cumxor == 0:
                ans = 0
            else:
                ans = 2
    
    print(ans)

if __name__=='__main__':
    main()