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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>

int main(void) {
  long long int n;
  scanf("%lld", &n);
  int d[n];
  int div = 1;
  for (long long int i = 0; i < n - 1; i++) {
    div *= 10;
  }

  int num;
  scanf(" %d", &num);
  for (long long int i = 0; i < n; i++) {
    d[i] = num / div;
    num %= div;
    div /= 10;
  }

  for (long long int i = 0; i < n - 1; i++) {
    for (long long int j = 0; j < n - i; j++) {
      d[j] = d[j] - d[j + 1];
      if (d[j] < 0) {
        d[j] *= -1;
      }
    }
  }

  printf("%d\n", d[0]);

  return 0;
}