#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e6+10;
int n,flag;
int fac[maxn];
int a[maxn];
void init(int n)
{
	fac[0]=0;
	for (int i=1;i<=n;i++)
	{
		fac[i]=fac[i-1];
		int tmp=i;
		while (!(tmp&1)) tmp/=2,fac[i]++;
	}
}
int C(int n,int m) {return (fac[n]-fac[m]-fac[n-m])==0;}
int main()
{
	cin>>n;getchar();
	for (int i=1;i<=n;i++)
	{
		char c=getchar();
		a[i]=c-48;
		if (a[i]==2) flag=1;
	}
	init(n);
	if (flag)
	{
		int tmp=0;
		for (int i=1;i<=n;i++)
			if (a[i]==2) tmp+=C(n-1,i-1);
		cout<<tmp%2<<endl;
		return 0;
	}
	int tmp=0;
	for (int i=1;i<=n;i++)
		if (a[i]==3) tmp+=C(n-1,i-1);
	cout<<tmp%2*2<<endl;
    return 0;
}

### Prompt

Your challenge is to write 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;
#define ll long long
const int maxn=1e6+10;
int n,flag;
int fac[maxn];
int a[maxn];
void init(int n)
{
	fac[0]=0;
	for (int i=1;i<=n;i++)
	{
		fac[i]=fac[i-1];
		int tmp=i;
		while (!(tmp&1)) tmp/=2,fac[i]++;
	}
}
int C(int n,int m) {return (fac[n]-fac[m]-fac[n-m])==0;}
int main()
{
	cin>>n;getchar();
	for (int i=1;i<=n;i++)
	{
		char c=getchar();
		a[i]=c-48;
		if (a[i]==2) flag=1;
	}
	init(n);
	if (flag)
	{
		int tmp=0;
		for (int i=1;i<=n;i++)
			if (a[i]==2) tmp+=C(n-1,i-1);
		cout<<tmp%2<<endl;
		return 0;
	}
	int tmp=0;
	for (int i=1;i<=n;i++)
		if (a[i]==3) tmp+=C(n-1,i-1);
	cout<<tmp%2*2<<endl;
    return 0;
}

```