#include <bits/stdc++.h>
#define forn(i, n) for(int i = 0; i < n; ++i)
using namespace std;

int main() {
	int n; string s;
	cin >> n >> s;

	vector<int> val(n, 0);
	forn(i, n-1) val[i+1] = val[i] + 32 - __builtin_clz((i+1) & (-i-1));

	int total = 0;
	forn(i, n) total ^= (s[i] - '1') * int(val[n-1] == val[n-i-1] + val[i]);
	if(s.find('2') != string::npos) total &= 1;
	cout << total << "\n";
}

### Prompt

Develop a solution in cpp to the problem described below:

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 <bits/stdc++.h>
#define forn(i, n) for(int i = 0; i < n; ++i)
using namespace std;

int main() {
	int n; string s;
	cin >> n >> s;

	vector<int> val(n, 0);
	forn(i, n-1) val[i+1] = val[i] + 32 - __builtin_clz((i+1) & (-i-1));

	int total = 0;
	forn(i, n) total ^= (s[i] - '1') * int(val[n-1] == val[n-i-1] + val[i]);
	if(s.find('2') != string::npos) total &= 1;
	cout << total << "\n";
}

```