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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int s[1000020];
int f(int a,int b){
  if(s[a]-s[b]-s[a-b-1]>0)return 1;
  return 0;
}
int main(){
  int n,i,h,d[1000010]={};
  scanf("%d",&n);
  for(i=2;i<1e6+10;i++)s[i]=s[i/2]+1;
  for(i=1;i<1e6+10;i++)s[i]=s[i]+s[i-1];
  for(i=0;i<n;i++)scanf("%1d",&d[i]);
  for(i=h=0;i<n;i++){
    d[i]--;
    if(d[i]==1)h=1;
  }
  if(h){
    for(i=h=0;i<n;i++){
      if(d[i]==1)h=(h+f(n-1,i))%2;
    }
    printf("%d\n",h);
  }
  else{
    for(i=h=0;i<n;i++){
      if(d[i]==2)h=(h+f(n-1,i))%2;
    }
    printf("%d\n",h*2);
  }
  return 0;
}