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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define N 1000005
int n,t1,t2,cnt[N];bool fl;char a[N];
int main()
{
	scanf("%d %s",&n,a);
	if(n==2) {printf("%d\n",abs(a[0]-a[1]));return 0;}
	for(int i=1;i<=n;++i) if(!(i%2)) cnt[i]=cnt[i/2]+1;
	for(int i=1;i<=n;++i) cnt[i]+=cnt[i-1];
	for(int i=0,x,t;i<n-1;++i)
	{
		x=abs(a[i]-a[i+1]);
		if(cnt[i]<cnt[n-2]-cnt[n-2-i]) t=0;else t=1;
		if(x==1) t1=(t1+t)%2,fl=1;if(x==2) t2=(t2+t)%2; 
	}if(t1) puts("1");else if(fl || (!fl && !t2)) puts("0");else puts("2");
	return 0;
}