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 main(void){
int n;
  scanf("%d",&n);
  char s[n];
  int a[n][n];
  scanf("%s",&s);
  for(int i=0;i<n;i++){
    a[0][i]=s[i]-'0';
   // printf("%d",a[0][i]);
  }
 // printf("\n");
  int i;
  for(int m=n;m>=2;m--){   //111
    int flag=0,c=0;
	  for(i=0;i<m-1;i++){				//222
	a[n-m+1][i]=abs(a[n-m][i]-a[n-m][i+1]);
        if(c>0){//33
          if(a[n-m+1][i]!=a[n-m+1][i-1])
		flag=1;
          
        }//33
      // printf("%d ",a[n-m+1][i]);
          c++;
      }//222
         // printf("\n");
    
       if(m==2){
          printf("%d",a[n-1][0]);
                return 0;
           }
                
        if(flag==0){
          printf("0");
          return 0;
        }
     
    
  }//111
  
//printf("%d",a[n-1][0]);
  return 0;
}