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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int Factrial( int n );

int main (void){
    int N;
    int i, Z, num, cur, pre, Y;

    scanf("%d", &Z );
    scanf("%d", &N );

    Z = Z - 1;
    for ( num = 0; Z > 0; Z-- ){
//        printf("%d %d\n", N, Z);
        for ( Y = Z; Y > 0; Y-- ){
            num += abs(N / Factrial(Y) - N % Factrial(Y) / Factrial(Y - 1)) * Factrial(Y - 1);
          	N = N % Factrial(Y);
          	printf("%d %d\n", num);
        }
        N = num;
        num = 0;
    }
    printf("%d\n", N);
    return 0;
}

// 引数によって階乗を求める
int Factrial( int n ){
    int i, num = 1;
    for ( i = 0; i < n; i++ ){
        num = num * 10;
    }
    return num;
}