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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define maxn 1000005

using namespace std;

int a[maxn],n,sum[maxn];
char s[maxn];

void init()
{
    for (int i=1;i<=1000000;i++)
    {
        int tmp=0,c=i;
        while (!(c&1))
        {
            c>>=1;
            tmp++;
        }
        sum[i]=sum[i-1]+tmp;
    }
}

bool Check(int n,int m)
{
    m=min(m,n-m);
    if (m==0) return 1;
    if (sum[m]==sum[n]-sum[n-m]) return 1;
    return 0;
}

int main()
{
    init();
    scanf("%d",&n);
    scanf("%s",s);
    int ans=0;
    for (int i=1;i<=n;i++) a[i]=s[i-1]-'0';
    for (int i=1;i<=n;i++)
    {
        if (i==1) continue;
        bool f=0;
        for (int j=1;j<=n-i+1;j++)
        {
            a[j]=abs(a[j]-a[j+1]);
            if (a[j]>1) f=1;
        }
        if (!f)
        {
            if (i==n) ans=a[1];
            else for (int j=1;j<=n-i+1;j++) if (Check(n-i,j-1)) ans^=a[j];
            break;
        }
    }
    printf("%d\n",ans);
    return 0;
}