#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> ii;

int main(){
	//freopen("i.txt","r",stdin);
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	
	int n; string s; cin >> n >> s;
	
	vector<int> v;
	for(int i = 1;i < n;i++){
		v.push_back(abs((int)(s[i] - s[i-1])));
	}
	
	reverse(v.begin(),v.end());
	bool hasOne = false;
	for(int x : v){
		if(x == 1) hasOne = true;
	}
	
	
	int dom = 1;
	if(hasOne) dom = 1;
	else dom = 2;
	
	for(int i = 0;i < (int) v.size();i++){
		if(hasOne){
			if(v[i] == 2) v[i] = 0;
		}
	}
	
	
	int ans = 0;
	for(int i = 0;i < (int) v.size();i++){
		if(v[i] != 0){
			if(((n-2) & i) == i){
				ans = 1 - ans;
			}
		}
	}
	
	cout << dom * ans;
	
	
}

### 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>
using namespace std;
typedef pair<int,int> ii;

int main(){
	//freopen("i.txt","r",stdin);
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	
	
	int n; string s; cin >> n >> s;
	
	vector<int> v;
	for(int i = 1;i < n;i++){
		v.push_back(abs((int)(s[i] - s[i-1])));
	}
	
	reverse(v.begin(),v.end());
	bool hasOne = false;
	for(int x : v){
		if(x == 1) hasOne = true;
	}
	
	
	int dom = 1;
	if(hasOne) dom = 1;
	else dom = 2;
	
	for(int i = 0;i < (int) v.size();i++){
		if(hasOne){
			if(v[i] == 2) v[i] = 0;
		}
	}
	
	
	int ans = 0;
	for(int i = 0;i < (int) v.size();i++){
		if(v[i] != 0){
			if(((n-2) & i) == i){
				ans = 1 - ans;
			}
		}
	}
	
	cout << dom * ans;
	
	
}

```