Problem:
Score : 700 points

Problem Statement

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}.

Sample Input 1

4
1231

Sample Output 1

1

x_{1,1},x_{1,2},x_{1,3},x_{1,4} are respectively 1,2,3,1.

x_{2,1},x_{2,2},x_{2,3} are respectively |1-2| = 1,|2-3| = 1,|3-1| = 2.

x_{3,1},x_{3,2} are respectively |1-1| = 0,|1-2| = 1.

Finally, x_{4,1} = |0-1| = 1, so the answer is 1.

Sample Input 2

10
2311312312

Sample Output 2

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#define For(i,A,B) for(i=(A);i<=(B);++i)
const int N=1000050;
int a[N],b[N],cnt;
inline int Abs(int x){return x<0?-x:x;}
inline void work(int x,int k){
    if(x)for(;!(x&1);x>>=1)cnt+=k;
}
inline bool check(int *a,int n){
    int i;
    bool f=0;
    cnt=0;
    For(i,1,n){
        if(!cnt&&(a[i]&1))f^=1;
        work(n-i,1);
        work(i,-1);
    }
    return f;
}
int main(){
    int n,i;
    bool f=0,g=0;
    scanf("%d",&n);
    For(i,1,n)scanf("%1d",a+i);
    if(check(a,n)){
        puts("1");
        return 0;
    }
    For(i,2,n){
        if(Abs(a[i]-a[i-1])==1){
            puts("0");
            return 0;
        }
        b[i-1]=Abs(a[i]-a[i-1])>>1;
    }
    puts(check(b,n-1)?"2":"0");
    return 0;
}