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>
using namespace std;

const int p = 5;
int A[1000009], B[1000009];

int f(int x) {
	return max(x, -x);
}

int main() {
	int N; scanf("%d", &N);
	for(int i=1; i<=N; i++) scanf("%1d", &A[i]);
	for(int i=1; i<=min(p, N-1); i++) {
		for(int j=1; j<=N-i; j++) A[j] = f(A[j] - A[j+1]);
	}
	if(N <= p+1) printf("%d", A[1]);
	else {
		N -= p;
		for(int i=1; i<=N; i++) B[i] = A[i];
		sort(B+1, B+N+1);
		if(B[N] == B[1]) puts("0");
		else if(B[N] == 2) {
			int c = 0;
			for(int i=1; i<=N; i++) if(A[i] == 2) {
				if(((N-i) & (i-1)) == 0) c ^= 1;
			}
			if(c) puts("2");
			else puts("0");
		}
		else {
			int c = 0;
			for(int i=1; i<=N; i++) if(A[i] == 1) {
				if(((N-i) & (i-1)) == 0) c ^= 1;
			}
			if(c) puts("1");
			else puts("0");
		}
	}
	return 0;
}