#include <iostream>
#include <vector>
#include <string>
using namespace std;

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int N;
	string S;
	cin >> N >> S;
	vector<char> A(N);
	for(int i = 0; i < N; i++) A[i] = S[i]-'1';
	vector<int> e2_fac(N+1, 0);
	for(int i = 1; i <= N; i++) e2_fac[i] = e2_fac[i/2] + i/2;
	char p = 0;
	for(int i = 0; i < N; i++) if(e2_fac[N-1] == e2_fac[i] + e2_fac[N-1-i]) p ^= A[i];
	if(p&1) {
		cout << "1\n";
		return 0;
	}
	bool has_odd = false;
	for(int i = 0; i < N; i++) if(A[i]&1) has_odd = true;
	if(has_odd) {
		cout << "0\n";
		return 0;
	}
	if(p&2) cout << "2\n";
	else cout << "0\n";
}

### Prompt

Please create a solution in CPP to the following problem:

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}.

Examples

Input

4
1231


Output

1


Input

10
2311312312


Output

0

### Response

```cpp
#include <iostream>
#include <vector>
#include <string>
using namespace std;

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int N;
	string S;
	cin >> N >> S;
	vector<char> A(N);
	for(int i = 0; i < N; i++) A[i] = S[i]-'1';
	vector<int> e2_fac(N+1, 0);
	for(int i = 1; i <= N; i++) e2_fac[i] = e2_fac[i/2] + i/2;
	char p = 0;
	for(int i = 0; i < N; i++) if(e2_fac[N-1] == e2_fac[i] + e2_fac[N-1-i]) p ^= A[i];
	if(p&1) {
		cout << "1\n";
		return 0;
	}
	bool has_odd = false;
	for(int i = 0; i < N; i++) if(A[i]&1) has_odd = true;
	if(has_odd) {
		cout << "0\n";
		return 0;
	}
	if(p&2) cout << "2\n";
	else cout << "0\n";
}

```