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 n;
int in[1000010];
int ne[1000010], cnt;
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    int last = 0;
    for (int i = 0; i < n; i++) {
        char a;
        cin >> a;
        in[i] = a - '0';
        if (last != in[i]) {
            in[cnt++] = in[i];
            last = in[i];
        }
    }
    if (cnt > 1e4) {
        cout << "0\n";
        return 0;
    }
    for (int i = cnt; i; i--) {
        for (int j = 0; j < i - 1; j++)
            in[j] = abs(in[j] - in[j + 1]);
    }
    cout << in[0] << '\n';
    return 0;
}