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 <stdio.h>
#include <queue>
#include <utility>
using namespace std;
typedef long long ll;

int main(void) {
  ll i, j, k, n;
  scanf("%lld", &n);
  char a[n + 10];
  bool b = true;
  scanf("%s", a);
  if(a[0] == 0) {
    for(i = 0; i < n - 1; ++i) if(a[i] != '1') b = false;
    if(b && a[n - 1] == '3') {
      printf("2");
      return 0;
    }
    b = true;
    for(i = 1; i < n; ++i) if(a[i] != '3') b = false;
    if(b) {
      printf("2");
      return 0;
    }
  } else if(a[0] == 2) {
    for(i = 0; i < n - 1; ++i) if(a[i] != '3') b = false;
    if(b && a[n - 1] == '3') {
      printf("2");
      return 0;
    }
    b = true;
    for(i = 1; i < n; ++i) if(a[i] != '1') b = false;
    if(b) {
      printf("2");
      return 0;
    }
  }
  if(n & 1) printf("%lld", (a[0] - a[n - 1]) & 1 ? 1 : 0);
  else {
    i = a[0] - a[1];
    j = a[n - 1] - a[n - 2];
    if(i < 0) i = -i;
    if(j < 0) j = -j;
    printf("%lld", (i - j) & 1 ? 1 : 0);
  }
  return 0;
}