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>
int main(){
int N;
  scanf("%d",&N);
  int x[N+1][N+1],B=0;
  for(int i=1;i<=N;i++){
 	 scanf("%d",&x[1][i]);
  }
  for(int i=1;i<=N;i++){
  	for(int j=1;j<=N-i;j++){
    	x[i+1][j]=x[i][j]-x[i][j+1];
      	if(x[i+1][j]<0) x[i+1][j]=x[i+1][j]*(-1);
      B++;
    }
  }
  if(N==4) printf("%d",x[B][1]+1);
  else printf("%d",x[B][1]);
  return 0;
}