#include <bits/stdc++.h>
using namespace std;

int a[1000005], f[1000005];

int main() {
	int N;
	cin >> N;
	for (int i = 0; i < N; ++i) {
	    char c;
	    cin >> c;
	    a[i] = c - '1';
	}

	f[0] = 0;
	for (int i = 0; i < N; ++i) f[i + 1] = __builtin_ctz(i + 1) + f[i];

	int ans = 0;
	for (int i = 0; i < N; ++i) {
		ans += a[i] * (f[N - 1] == f[i] + f[N - i - 1]);
	}
	if (ans % 2) {
		cout << 1 << '\n';
		return 0;
	}
	for (int i = 0; i < N; ++i) {
		if (a[i] == 1) {
			cout << 0 << '\n';
			return 0;
		}	
	}
	ans = 0;
	for (int i = 0; i < N; ++i) {
		ans += (a[i] / 2) * (f[N - 1] == f[i] + f[N - i - 1]);
	}
	cout << 2 * (ans % 2) << '\n';
}
### Prompt

Your task is to create a cpp solution 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 <bits/stdc++.h>
using namespace std;

int a[1000005], f[1000005];

int main() {
	int N;
	cin >> N;
	for (int i = 0; i < N; ++i) {
	    char c;
	    cin >> c;
	    a[i] = c - '1';
	}

	f[0] = 0;
	for (int i = 0; i < N; ++i) f[i + 1] = __builtin_ctz(i + 1) + f[i];

	int ans = 0;
	for (int i = 0; i < N; ++i) {
		ans += a[i] * (f[N - 1] == f[i] + f[N - i - 1]);
	}
	if (ans % 2) {
		cout << 1 << '\n';
		return 0;
	}
	for (int i = 0; i < N; ++i) {
		if (a[i] == 1) {
			cout << 0 << '\n';
			return 0;
		}	
	}
	ans = 0;
	for (int i = 0; i < N; ++i) {
		ans += (a[i] / 2) * (f[N - 1] == f[i] + f[N - i - 1]);
	}
	cout << 2 * (ans % 2) << '\n';
}
```