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: Accepted)

#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <set>
#include <map>
#include <cassert>
#include <numeric>
#include <string>
#include <cstring>
#include <cmath>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long int int64;



const int N = (int) 1e6 + 100;
int a[3][N];

int getC(int n, int i)
{
    int j = 20;
    while (n >= 0)
    {
        while ((1 << j) > n)
            j--;
        if (i > n)
            return 0;
        //eprintf("(%d, %d) : -= %d\n", n, i, 1 << j);
        n -= (1 << j);
        if (i >= (1 << j))
            i -= (1 << j);
    }
    return 1;
}

char s[N];

int main(int,  char **)
{
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif

    for (int i = 0; i < 20; i++)
    {
        for (int j = 0; j < 20; j++)
            eprintf("%d", getC(i, j));
        printf("\n");
    }

    int n;
    scanf("%d", &n);
    scanf("%s", s);
    for (int i = 0; i < n; i++)
        a[0][i] = s[i] - '0';

    for (int t = 0; t < 2; t++)
    {
        for (int i = 0; i < n; i++)
        {
            a[t + 1][i] = abs(a[t][i] - a[t][i + 1]);
        }
    }

    if (n <= 3)
    {
        printf("%d\n", a[n - 1][0]);
        return 0;
    }

    int ans = 0;
    n -= 2;
    //eprintf("n = %d\n", n);
    int x = 2;

    for (int i = 0; i < n; i++)
        if (a[2][i] == 1)
            x = 1;
    for (int i = 0; i < n; i++)
    {
        //eprintf("%d", a[2][i]);
        ans ^= getC(n - 1, i) * (a[2][i] == x) * x;
    }
    eprintf("\n");

    printf("%d\n", ans);

	return 0;
}