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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main ()
{
    char x[1000000];
    int a,b,i,j;
    scanf("%d",&a);
    scanf("%s",x);
    b=a;
    for(i=0;i<a;i++)
    {
        x[i]=x[i]-'0';
    }
    for(i=0;i<a;i++)
    {
        for(j=0;j<b;j++)
        {
            x[j]= abs(x[j]-x[j+1]);
        }
        b=b-1;
        x[j]='\0'-'0';
    }
    for(i=0;i<a;i++)
    {
        x[i]=x[i]+'0';
    }
    printf("%s\n",x);
    return 0;
}