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;

int f(vector<int> X) {
  if ((int)X.size() == 0) return 0;
  if ((int)X.size() == 1) return X[0];
  vector<int> res;
  int n = X[0];
  for (int i = 1; i < (int)X.size(); i++) {
    if (n != X[i]) {
      res.emplace_back(abs(n - X[i]));
      n = X[i];
    }
  }
  return f(res);
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  string S;
  cin >> S;
  vector<int> A;
  for (int i = 0; i < N; i++) {
    A.emplace_back(S[i] - '0');
  }

  /* function<int(vector<int>)> f = [&](vector<int> X) { */
  /*   /1* for (auto &v : X) { *1/ */
  /*   /1*   cout << v << (&v == &X.back() ? '\n' : ' '); *1/ */
  /*   /1* } *1/ */
  /*   if ((int)X.size() == 0) return 0; */
  /*   if ((int)X.size() == 1) return X[0]; */
  /*   vector<int> res; */
  /*   int n = X[0]; */
  /*   for (int i = 1; i < (int)X.size(); i++) { */
  /*     if (n != X[i]) { */
  /*       res.emplace_back(abs(n - X[i])); */
  /*       n = X[i]; */
  /*     } */
  /*   } */
  /*   return f(res); */
  /* }; */

  cout << f(A) << '\n';
  return 0;
}