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 <bits/stdc++.h>
using namespace std;
 
int main() {
  long long N;
  cin >> N;
  long long Q;
  cin >> Q;
  vector<long long> veca(N);
  for(long long j=0;j<N;j++){
    veca.at(N-j-1)=Q%10;
      Q=(Q-Q%10)/10;
  }
  vector<vector<int>> vecx(N,vector<int>(N));
  for(long long i=0;i<N;i++){
    for(long long j=0;j<N-i;j++){
      if(i!=0)
      vecx.at(i).at(j)=max(vecx.at(i-1).at(j)-vecx.at(i-1).at(j+1),vecx.at(i-1).at(j+1)-vecx.at(i-1).at(j));
      else
        vecx.at(i).at(j) = veca.at(j);
    }
  }
  cout << vecx.at(N-1).at(0) << endl;
}