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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007
//#define P 2

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}
int compw(const void *a, const void *b){return (*(int*)a>*(int*)b)-(*(int*)a<*(int*)b);}

uint32_t nextpint(void){ char c=getchar(); while(c<'0'||'9'<c) c=getchar(); uint_fast32_t x=0; while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); } return x; }

inline int min2(const int a, const int b){return a>b?b:a;}
inline int dif2(const int a, const int b){return a>b?a-b:b-a;}

int k[1000000];
int rk[1000000];

int getk(int a, int x){
	if(x==1) return a;
	int b = getk(a, x/2);
	b = ((long)b)*b%P;
	if((x&1)==0){
		return b;
	}else{
		return ((long)b)*a%P;
	}
}
 
int comb(int a, int b){
	if(a<b) return 0;
	return ((long)k[a])*rk[b]%P*rk[a-b]%P;
}

void init(int n){
	k[0]=1;
	for(int i=1; i<=n; i++){
		k[i]=k[i-1]*((long)i)%P;
	}	
	rk[n]=getk(k[n], P-2);
	for(int i=n-1; i>=0; i--){
		rk[i]=((long)rk[i+1])*(i+1)%P;
	}
}

char a[1000000];
int main(void){
	int n=nextpint();
	int f=1;
	for(int i=0; i<n; i++){
		a[i]=getchar()-'0';
		if(a[i]==1) f=0;
	}
	if(f!=0){
		for(int i=0; i<n; i++){
			a[i]>>=1;
		}
	}
	init(n-1);
	int x=0;
	for(int i=0; i<n; i++){
		x+=comb(n-1,i)*a[i];
		x&=1;
	}
	if(f==0){
		putchar(x+'0');
	}else{
		putchar((x<<1)+'0');
	}
	putchar(10);
}