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

const int NMAX = 1e6 + 5;

int n;
int a[NMAX];
int b[NMAX];
bool am;

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++) {
        char x; cin >> x;
        a[i] = x - '1';
        if (a[i] == 1)
            am = 1;
        b[a[i]] += (((n - 1) & i) == i);
    }

    if (b[1] % 2 == 1) // imi da impar
        cout << 1;
    else if (!am && (b[2] % 2 == 1)) // nu am 1 si nici n-o sa am
        cout << 2;
    else // am 1
        cout << 0;


    return 0;
}

### Prompt

Your task is to create 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;

const int NMAX = 1e6 + 5;

int n;
int a[NMAX];
int b[NMAX];
bool am;

int main()
{
    cin >> n;
    for (int i = 0; i < n; i++) {
        char x; cin >> x;
        a[i] = x - '1';
        if (a[i] == 1)
            am = 1;
        b[a[i]] += (((n - 1) & i) == i);
    }

    if (b[1] % 2 == 1) // imi da impar
        cout << 1;
    else if (!am && (b[2] % 2 == 1)) // nu am 1 si nici n-o sa am
        cout << 2;
    else // am 1
        cout << 0;


    return 0;
}

```