#include<bits/stdc++.h>
#define debug1 cout<<"?"<<endl;
#define debug2 cout<<"?"<<i<<endl;

typedef long long ll;
#define PII pair<ll,ll>
const int maxn=1e6+9;
const double pi=acos(-1);
const int mod=998244353;
const int INF=0x3f3f3f3f;
const double ep=1e-10;
using namespace std;
int main()
{
	int n,i,a[maxn]={0},vis[4]={0};
	char s[maxn];
	scanf("%d",&n);
	scanf("%s",s+1);
	for(i=1;i<=n;i++)a[i]=s[i]-'0'-1,vis[a[i]]=1;
	if(vis[2])
	{
		if(vis[1])
		{
			for(i=1;i<=n;i++)a[i]=a[i]%2;
		}
		else for(i=1;i<=n;i++)if(a[i]==2)a[i]=1;
	}
	int ans=0;
	for(i=0;i<n;i++)
	{
		if(((n-1)&i)==i)ans=(ans+a[i+1])%2;
	}
	if(ans)
	{
		if(vis[1])puts("1");
		else puts("2");
	}
	else puts("0");
}
### Prompt

In cpp, your task is to solve 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>
#define debug1 cout<<"?"<<endl;
#define debug2 cout<<"?"<<i<<endl;

typedef long long ll;
#define PII pair<ll,ll>
const int maxn=1e6+9;
const double pi=acos(-1);
const int mod=998244353;
const int INF=0x3f3f3f3f;
const double ep=1e-10;
using namespace std;
int main()
{
	int n,i,a[maxn]={0},vis[4]={0};
	char s[maxn];
	scanf("%d",&n);
	scanf("%s",s+1);
	for(i=1;i<=n;i++)a[i]=s[i]-'0'-1,vis[a[i]]=1;
	if(vis[2])
	{
		if(vis[1])
		{
			for(i=1;i<=n;i++)a[i]=a[i]%2;
		}
		else for(i=1;i<=n;i++)if(a[i]==2)a[i]=1;
	}
	int ans=0;
	for(i=0;i<n;i++)
	{
		if(((n-1)&i)==i)ans=(ans+a[i+1])%2;
	}
	if(ans)
	{
		if(vis[1])puts("1");
		else puts("2");
	}
	else puts("0");
}
```