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;

const int MAXN = 1e6 + 5;
int x[100][MAXN];

int c(int n, int k) {
    if (n == 0) return 1;
    if (n == 1) return 1;
    if (n == 2) return k != 1;
    if (n == 3) return 1;
    return c(n / 2, k / 2) * c(n % 2, k % 2);
}

signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        char c;
        cin >> c;
        x[1][i] = c - '0';
    }
    for (int k = 2; k <= 100; k++) {
        for (int i = 1; i < n - k + 1; i++)
            x[2][i] = abs(x[1][i] - x[1][i+1]);
        if (n == k) {
            cout << x[k][1] << endl;
            return 0;
        }
    }
    for (int i = 1; i < n - 4; i++)
        x[6][i] = abs(x[5][i] - x[5][i+1]);
    int m = n - 100;
    int ans = 0;
    for (int i = 1; i < n - 100 + 1; i++) {
        ans ^= c(m + 1, i) * x[100][i];
    }
    cout << ans << endl;
}