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

int main(){
	int N, i;
	scanf("%d", &N);
	vector<int> a(N);
	for(i = 0; i < N; i++){
		scanf("%1d", &a[i]);
	}
	int b = 0;
	for(i = 0; i < N; i++){
		b ^= ((((N - 1) & i) == i) * (a[i] & 1));
	}
	if(b == 1){
		printf("1\n");
		return 0;
	}
	vector<int> a2(N - 1);
	for(i = 0; i < N - 1; i++){
		a2[i] = abs(a[i] - a[i + 1]);
		if(a2[i] == 1){
			printf("0\n");
			return 0;
		}
	}
	b = 0;
	for(i = 0; i < N - 1; i++){
		b ^= ((((N - 2) & i) == i) * (a2[i] == 2));
	}
	if(b == 1){
		printf("2\n");
	}
	else{
		printf("0\n");
	}
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 main(){
	int N, i;
	scanf("%d", &N);
	vector<int> a(N);
	for(i = 0; i < N; i++){
		scanf("%1d", &a[i]);
	}
	int b = 0;
	for(i = 0; i < N; i++){
		b ^= ((((N - 1) & i) == i) * (a[i] & 1));
	}
	if(b == 1){
		printf("1\n");
		return 0;
	}
	vector<int> a2(N - 1);
	for(i = 0; i < N - 1; i++){
		a2[i] = abs(a[i] - a[i + 1]);
		if(a2[i] == 1){
			printf("0\n");
			return 0;
		}
	}
	b = 0;
	for(i = 0; i < N - 1; i++){
		b ^= ((((N - 2) & i) == i) * (a2[i] == 2));
	}
	if(b == 1){
		printf("2\n");
	}
	else{
		printf("0\n");
	}
	return 0;
}
```