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

Solution (in JAVA):

Solution:
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = Integer.parseInt(scanner.nextLine());
        if(n==1){
            System.out.println(scanner.nextInt());
            return;
        }
        char[] chars = scanner.nextLine().toCharArray();
        int[] a = new int[n-1];
        for (int i = 1; i < n; i++) {
            a[i-1] = Math.abs(chars[i]-chars[i-1]);
        }
        if(n==2){
            System.out.println(a[0]);
            return;
        };
        System.out.println(solution(a));
    }

    public static int solution(int[] a){
        int n = a.length;
        if(!findParity(a)) return 1;
        for (int i = 0; i < n; i++) {
            if(a[i]==1) return 0;
            a[i]/=2;
        }
        if(findParity(a)) return 0;
        else return 2;
    }

    //true - четнео, афдыу - нечетно
    private static boolean findParity(int[] a){
        int n = a.length;
        int ans = a[0]^a[n-1];
        int pow = 0;
        for (int i = 1; i < n-1; i++) {
            if((i&1)==0){
                int c = 0;
                int t = i;
                while ((t&1)==0){
                    t>>=1;
                    c++;
                }
                pow-=c;
            }
            if(((n-i)&1)==0){
                int c = 0;
                int t = n-i;
                while ((t&1)==0){
                    t>>=1;
                    c++;
                }
                pow+=c;
            }
            if(pow==0){
                ans = ans ^ a[i];
            }
        }
        return (ans&1)==0;
    }
}