#include<bits/stdc++.h>
using namespace std;
int n;
int a[1000100];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%1d",&a[i]);
    if(n==1)
    {
        printf("%d\n",a[1]);
        return 0;
    }
    n--;
    for(int i=1;i<=n;++i)a[i]=abs(a[i]-a[i+1]);
    int eo=0;
    int fl=0;
    for(int i=1;i<=n;++i)
    {
        eo^=(a[i]&1)*(((n-1)&(i-1))==(i-1));
        if(a[i]==1)fl=1;
    }
    if(eo)
    {
        puts("1");
        return 0;
    }
    if(fl)
    {
        puts("0");
        return 0;
    }
    for(int i=1;i<=n;++i)a[i]/=2;
    eo=0;
    for(int i=1;i<=n;++i)
        eo^=(a[i]&1)*(((n-1)&(i-1))==(i-1));
    puts(eo?"2":"0");
    return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int n;
int a[1000100];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%1d",&a[i]);
    if(n==1)
    {
        printf("%d\n",a[1]);
        return 0;
    }
    n--;
    for(int i=1;i<=n;++i)a[i]=abs(a[i]-a[i+1]);
    int eo=0;
    int fl=0;
    for(int i=1;i<=n;++i)
    {
        eo^=(a[i]&1)*(((n-1)&(i-1))==(i-1));
        if(a[i]==1)fl=1;
    }
    if(eo)
    {
        puts("1");
        return 0;
    }
    if(fl)
    {
        puts("0");
        return 0;
    }
    for(int i=1;i<=n;++i)a[i]/=2;
    eo=0;
    for(int i=1;i<=n;++i)
        eo^=(a[i]&1)*(((n-1)&(i-1))==(i-1));
    puts(eo?"2":"0");
    return 0;
}

```