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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
int n,a[1000005],b[1000005];
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		char x;
		cin>>x;
		a[i]=x-'0';
	}
	while(n>1){
		bool flag=0;
		for(int i=2;i<=n;i++)if(a[i]!=a[1])flag=1;
		if(!flag)return puts("0"),0;
		int tot=0;
		for(int i=1;i<=n;i++){
			int j=i+1;
			while(j<=n&&a[i]==a[j])j++;
			if(j-i>=2){
				b[++tot]=a[i],b[++tot]=a[i];
				i=j-1;
				continue; 
			}
			b[++tot]=a[i];
		}
		for(int i=1;i<tot;i++)a[i]=abs(b[i]-b[i+1]);
		n=tot-1;
	}
	cout<<a[1];
	return 0;
}