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: Wrong Answer)

#include "bits/stdc++.h"
#define YES "YES"
#define NO "NO"
#define Yes "Yes"
#define No "No"
#define DEFAULT LL ans=solve();\
 	if(ans==NONE){\
 	}else{\
 		cout << ans << endl;\
 	}
#define YESNO three(solve(),(OUT(YES),1),(OUT(NO),0))
#define YesNo three(solve(),OUT(Yes),OUT(No))
#define ECHO OUT(solve())
#define three(A,B,C) ((A)?(B):(C))
#define FOR(i,a,b)  for(LL i=(a);i< (LL)(b);i++)
#define EFOR(i,a,b) for(LL i=(a);i<=(LL)(b);i++)
#define RFOR(i,a,b) for(LL i=(b-1);i>=(LL)(a);i--)
#define REP(i,b) FOR(i,zero,b)
#define rep REP
#define EREP(i,b) EFOR(i,zero,b)
#define RREP(i,b) RFOR(i,zero,b)
#define ALL(c) c.begin(),c.end()
#define UNIQUE(c) sort(ALL(c));c.erase(unique(ALL(c)),c.end())
#define MAX(c) (*max_element(ALL(c)))
#define MIN(c) (*min_element(ALL(c)))
#define MP make_pair
#define FI first
#define SE second
#define SI(x) (LL(x.size()))
#define PB push_back
#define DEBUG(a) OUT(a)
#define DEBUG2(a,b) OUT2(a,b)
#define cat cout << __LINE__ << endl
#define OUT(a) cout << (a) << endl
#define OUT2(a,b) cout << (a) <<" "<<(b) << endl
#define zero 0LL
#define all ALL
#define pb emplace_back
#define eb pb
#define int long long
using namespace std;
template<typename T> inline void maximize(T &a, T b) { a = max(a, b); }
template<typename T> inline void minimize(T &a, T b) { a = min(a, b); }
template<typename T> inline bool middle(T a, T b, T c) { return b <= a && a <= c; }
template<class T> inline bool MX(T &l, const T &r) { return l < r ? l = r, 1 : 0; }
template<class T> inline bool MN(T &l, const T &r) { return l > r ? l = r, 1 : 0; }
typedef int LL;
typedef double ld;
typedef int ut;
typedef vector<ut> VI;
typedef vector<VI> VII;
typedef pair<ut, ut> pr;
typedef pair<ut, pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef tuple<int, int, int, int> tapu;
typedef vector<tapu> Vtapu;
typedef priority_queue<pr, Vpr > PQ;
inline void outputVI(VI x) { REP(i, SI(x)) { cout << three(i, " ", "") << x[i]; }OUT(""); }
const int SIZE1 = 5e5 + 1000;
const int SIZE2 = 5010;
const int SIZE3 = 430;
const int SIZE = SIZE1;
const int MAPSIZE = 40;
const LL p998 = 998244353;
const LL p107 = 1000000007;
const LL p = p107;
const LL INF = 1LL << 60;
const long double EPS = 1e-7;
const LL NONE=-2;
#define endl "\n"
ut N, M,K,  Q, D, H, W;
VI edges[SIZE];
LL vals[SIZE], answer = zero;
LL A[SIZE],B[SIZE],C[SIZE];
LL counts[3];
LL num(int n){
	LL ans=0;
	while(n){
		ans+=n/2;
		n/=2;
	}
	return ans;
}

LL nCr(int n,int r){

	return num(n)-num(r)-num(n-r)>=1;
}
bool exist[SIZE];
LL solve(){
	cin >> N;
	string s;
	cin >> s;
	REP(i,N){
		A[i]=s[i]-'0';
	}
	N--;
	REP(i,N){
		B[i]=abs(A[i]-A[i+1]);
	}
	REP(i,N){
		counts[B[i]]+=nCr(N-1,i);	
		exist[B[i]]|=1;
	}
	if(counts[1]) return 1;
	if(counts[2]%2 and exist[1]==0) return 2;
	return 0;
	return NONE;
}
 signed main() {
 
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(10);
	LL T=1;

	REP(i,T){
		DEFAULT;

	}
	
//	YesNo;
//	YESNO;
//	three(solve(),OUT("First"),OUT("Second"));
	//cin >> N;
	return 0;
 
}