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: Runtime Error)

#include <stdio.h>

int main() {
	int N;
	int a[100000];
	int i, j;
	int x;
	char c;
	scanf("%d", &N);
	getchar();
	for (i = 1; i <= N; i++) {
		scanf("%c", &c);
		a[i] = c - '0';
	}

	for (i = 0; i < N - 1; i++) {
		a[0] = a[1];
		for (j = N - 1; j >= 1; j--) {
			x = a[N - j + 1] - a[0];
			a[0] = a[N - j];
			if (x < 0)a[N - j] = -x;
			else a[N - j] = x;

		}
	}
	printf("%d", a[1]);
	return 0;

}