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: Time Limit Exceeded)

#include <bits/stdc++.h>
//#pragma GCC optimize("Ofast")

#define fi first
#define se second

const int N = 110;

using namespace std;

void add(vector < pair < int, int > > &v, pair < int, int > p)
{
        if(v.empty() || v.back().fi != p.fi) v.push_back(p);
        else v.back().se += p.se;
}

int main()
{
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);

        ios_base::sync_with_stdio(0);

        int n;
        cin >> n;
        vector < pair < int, int > > v, nv;
        for(int i = 1; i <= n; i++){
                char c;
                cin >> c;
                int x = c - '0';
                add(v, make_pair(x, 1));
        }
        for(int i = 1; i < n; i++){
                nv.clear();
                int last = -1;
                for(auto p: v){
                        if(last != -1) add(nv, make_pair(abs(last - p.fi), 1));
                        if(p.se > 1) add(nv, make_pair(0, p.se - 1));
                        last = p.fi;
                }
                v = nv;
        }

        cout << v.back().fi << "\n";
}