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>
#ifdef LOCAL_DEBUG
#define DEBUG 1
#define CERR if(DEBUG) cerr
#define MACRO(_1,_2,_3,_4,_5,_6,_7,_8,_9,_10,NAME,...) NAME
#define pr(...) CERR<<MACRO(__VA_ARGS__,pr10,pr9,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__)<<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
#define pr9(a,b,c,d,e,f,g,h,i) pr1(a)<<pr8(b,c,d,e,f,g,h,i)
#define pr10(a,b,c,d,e,f,g,h,i,j) pr1(a)<<pr9(b,c,d,e,f,g,h,i,j)
#define prArr(a){CERR<<(#a)<<"={";int i=0;for(auto t:(a))CERR<<(i++?", ":"")<<t;CERR<<"}"<<endl;}
#else
#define DEBUG 0
#define pr(...)
#define prArr(a)
#endif
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<60)+1e9; // ~ 1.15 * 1e18
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}

template<typename T>
class CumulativeSum{
public:
  Int n;
  vector<T> sum;
  vector<T> A;
  Int added;
  CumulativeSum():n(-1),added(0){}
  CumulativeSum(Int n):n(n), sum(n+1), A(n+1),added(0){}
  CumulativeSum(const vector<T> &B):n(B.size()), sum(n+1), A(n+1),added(0){
    for(Int i=1;i<=n;i++) sum[i] = sum[i-1] + B[i-1];
  }

  void apply(){
    for(Int i=1;i<=n;i++) A[i] = A[i] + A[i-1];
    for(Int i=1;i<=n;i++) A[i] = A[i] + A[i-1];
    for(Int i=1;i<=n;i++) sum[i] = sum[i] + A[i-1];
    added = 0; A.clear(); A.resize(n+1);
  }

  //[l, r)にxを加算
  void add(Int l, Int r, T x){
    added = 1;
    assert(l <= r && 0 <= l && r <= n);
    A[l] = A[l] + x;
    A[r] = A[r] - x;
  }

  //[l, r)の和を得る
  T get(Int l,Int r){
    assert(l<=r && 0<=l && r<=n);
    if(added) apply();
    return sum[r] - sum[l];
  }
};

//INSERT ABOVE HERE
// 0 0 -> 0
// 1 1 -> 0
// 2 2 -> 0

// 0 1 -> 1
// 1 0 -> 1
// 1 2 -> 1
// 2 1 -> 1

// 0 2 -> 2
// 2 0 -> 2

CumulativeSum<Int> S;
Int nCr(Int n, Int r){
  if(S.n == -1){
    S = CumulativeSum<Int>(2e6);
    for(Int i=0;i<2e6;i++){
      Int c = 0;
      Int t = i;
      while(t && t%2 == 0) t/=2, c++;
      S.add(i, i+1, c);
    }
  }
  if(n < r) return 0;
  if(r == 0) return 1;
  if(n == r) return 1;
  Int a = S.get(0, r+1); //r!
  Int b = S.get(n - r + 1, n + 1); // (n!) / (n-r)!
  //pr(n, r, a, b);
  if(b > a) return 0;
  return 1;
}
Int solve_(Int N, vector<Int> A, Int flag = 0){

  for(Int i=0;i<N-1;i++){
    if(i == 0 && flag) pr(i, A);
    vector<Int> nA;
    for(Int i=0;i<(Int)A.size()-1;i++){
      nA.push_back(abs(A[i] - A[i+1]));
    }
    A = nA;
    if(flag)pr(i, A);
  }
  assert(A.size() == 1);
  return A[0];
}

Int solve2(Int N, vector<Int> A, Int flag){
  Int sum = 0;
  for(Int i=0;i<(Int)A.size();i++){
    Int a = A[i];
    Int num =nCr(((Int)A.size()-1), i);
    pr(a, num, A.size(), i);
    if(a == 2) sum ^= num % 2;
  }
  pr(sum);
  return sum? 2:0;
}

Int solve(Int N, vector<Int> A, Int flag = 0){
  Int all02 = 1;
  for(Int i=0;i<N;i++) if(A[i] == 1) all02 = 0;
  if(all02) return solve2(N, A, flag);

  for(Int &a:A) a %= 2;
  for(Int i=0;i<N-1;i++){
    vector<Int> nA;
    Int flag = 1;
    for(Int i=0;i<(Int)A.size()-1;i++){
      nA.push_back(abs(A[i] - A[i+1]));
      if(nA.back() == 2) flag = 0;
    }
    A = nA;
    if(flag) break;
  }
  if(A.size() == 1) return A[0];
  if(flag)pr(A);
  Int sum = 0;
  for(Int i=0;i<(Int)A.size();i++){
    Int a = A[i];
    Int num =nCr(((Int)A.size()-1), i);
    if(flag) pr(a, num, A.size(), i);
    sum ^= a * num % 2;
  }
  return sum;
}

void test(){
  while(1){
    Int N = rand()%100 + 1;
    vector<Int> A(N);
    for(Int &a:A) a = rand()%3;
    Int a = solve_(N, A);
    //if(a == 2) continue;
    Int b = solve(N, A);
    //if(a == 2) pr(N, A, a, b), solve_(N, A, 1);

    assert(a == b);
    //assert(a != 2);
    //return;
  }
}

signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);
  //test();

  Int  N;
  cin>>N;
  string s;
  cin>>s;
  N = s.size();

  vector<Int> A;
  for(Int i=0;i<N;i++) A.push_back(s[i] - '0');
  Int ans = solve(N, A);
  cout<<ans<<endl;
  return 0;
}