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: Wrong Answer)

import math
import numpy as np

N = int(input())
arr_rev = list(map(int, list(input())))
arr_rev.reverse()
arr_rev = np.array(arr_rev)

def calc(a, b):
    return abs(-a + b)


def chanto(arr_rev):
    inp = arr_rev
    diff_1 = np.array([calc(inp[x], inp[x + 1]) for x in range(len(inp) - 1)])
    if diff_1[0] != 0 and diff_1[len(diff_1)-1] != 0:

        diff_1_nonz = diff_1[diff_1 > 0]

        if len(diff_1_nonz) == 0: # all zero
            return 0
        if len(diff_1_nonz) == 1: # only single value
            return diff_1_nonz[0]

        return (diff_1_nonz[0] + diff_1_nonz[len(diff_1_nonz) - 1]) % 2
    else:
        tmp = diff_1[diff_1 > 0]
        if np.all(tmp == 2):
            return 2
        else:
            if len(tmp[tmp==1]) % 2 == 1:
                return 1
            else:
                return 0


print(chanto(arr_rev))