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: Accepted)

// B - 123 Triangle
#include <bits/stdc++.h>

using namespace std;

int main(int argc, char *argv[])
{
	int n;
	string s;
	cin >> n >> s;
	vector<int> a(n);
	for (int i = 0; i < n; i++) a[i] = s[i] - '1';
	vector<int> c(n);
	for (int i = 1; i < c.size(); i++) c[i] = __builtin_ctz(i) + c[i - 1];
	int ans = 0;
	for (int i = 0; i < n; i++) {
		int d = c[n - 1] - c[i] - c[n - 1 - i];
		ans ^= d ? 0 : (a[i] % 2);
	}

	if (ans != 0) {
		cout << 1 << endl;
	} else {
		bool has_one = false;
		for (int i = 0; i < a.size(); i++) if (a[i] == 1) has_one = true;
		if (has_one) {
			cout << 0 << endl;
		} else {
			for (int i = 0; i < a.size(); i++) a[i] /= 2;
			int ans = 0;
			for (int i = 0; i < n; i++) {
				int d = c[n - 1] - c[i] - c[n - 1 - i];
				ans ^= d ? 0 : (a[i] % 2);
			}
			cout << (ans ? 2 : 0) << endl;
		}
	}

	return 0;
}