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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <vector>
int a;
char x[1000010];
std::vector<int> V,V2;
int abs(int k)
{
	return k>0?k:-k;
}
int comb[1000010];

int main()
{
	int a;
	scanf("%d",&a);
	scanf("%s",x+1);
	for(int i=1;i<=a;i++) V.push_back(x[i]-'0');
	if(a<=50)
	{
		for(int i=1;i<a;i++)
		{
			V2.clear();
			for(int j=1;j<V.size();j++) V2.push_back(abs(V[j]-V[j-1]));
			V.clear();
			V = V2;
		}
		printf("%d",V[0]);
		return 0;
	}
	
	for(int i=1;i<=30;i++)
	{
		V2.clear();
		for(int j=1;j<V.size();j++) V2.push_back(abs(V[j]-V[j-1]));
		V.clear();
		V = V2;
	}
	
	
	for(int i=0;i<V.size();i++)
	{
		int c = V.size()-1;
		while(c)
		{
			comb[i] += (c/2);
			c/=2;
		}
		c = i;
		while(c)
		{
			comb[i] -= (c/2);
			c/=2;
		}
		c = V.size()-1-i;
		while(c)
		{
			comb[i] -= (c/2);
			c/=2;
		}
	}
	int ans = 0;
	for(int i=0;i<V.size();i++)
	{
		if(comb[i]==0)
		{
			ans += V[i];
			ans %=2;
		}
	}
	printf("%d",ans);
}