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<bits/stdc++.h>
#include<cstdio>
#include<iostream>
#include<map>
#include<string>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<cstring>
using namespace std;
typedef long long ll;
ll a[1000005];
char s[1000005];
int main()
{
	ll n;
	scanf("%lld",&n);
	scanf(" %s",s);
	for(ll i=0;i<n;i++)
	{
		a[i]=s[i]-'0';
	}
	if(n==1)
	{
		printf("%lld",a[0]);
		return 0;
	}
	for(ll i=0;i<n-1;i++)
	{
		a[i]=abs(a[i+1]-a[i]);
	}
	n=n-1;
	ll a0=0,a1=0,a2=0;
	for(ll i=0;i<n;i++)
	{
		if(a[i]==0)			a0++;
		else if(a[i]==1)	a1++;
		else				a2++;
	}
	if(a1==0)
	{
		if(a[n-1]!=a[0])
		{
			printf("2");
		}
		else
		{
			printf("0");
		}
		return 0;
	}
	if(n==1)
	{
		printf("%lld",a[0]);
	}
	else
	{
		ll c=a[n-1]-a[0];
		if(c%2==0)
		{
			printf("0");
		}
		else
		{
			printf("1");
		}
	}
}