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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#ifdef ENABLE_DEBUG
#define dump(a) cerr<<#a<<"="<<a<<endl
#define dumparr(a,n) cerr<<#a<<"["<<n<<"]="<<a[n]<<endl
#else
#define dump(a) 
#define dumparr(a,n) 
#endif
#define FOR(i, a, b) for(ll i = (ll)a;i < (ll)b;i++)
#define For(i, a) FOR(i, 0, a)
#define REV(i, a, b) for(ll i = (ll)b-1LL;i >= (ll)a;i--)
#define Rev(i, a) REV(i, 0, a)
#define REP(a) For(i, a)
#define SIGN(a) (a==0?0:(a>0?1:-1))

typedef long long int ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef pair<ll, ll> pll;
typedef pair<ll,pll> ppll;
typedef vector<ll> vll;
typedef long double ld;
typedef pair<ld,ld> pdd;

const ll INF=(1LL<<50);
#if __cplusplus<201700L
ll gcd(ll a, ll b) {
  if(a < b) return gcd(b, a);
  ll r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
#endif
template<class T>
bool chmax(T& a,const T& b){
  if(a<b){
    a=b;
    return true;
  }
  return false;
}
template<class T>
bool chmin(T& a,const T& b){
  if(a>b){
    a=b;
    return true;
  }
  return false;
}
template<class S,class T>
std::ostream& operator<<(std::ostream& os,pair<S,T> a){
  os << "(" << a.first << "," << a.second << ")";
  return os;
}
template<class T>
std::ostream& operator<<(std::ostream& os,vector<T> a){
  os << "[ ";
  REP(a.size()){
    os<< a[i] << " ";
  }
  os<< " ]";
  return os;
}


list<ll> ans;
void f(list<ll>::iterator l){
  auto itr=ans.begin();
  while(next(itr)!=l){
    if(*itr<0){
      auto itr2=next(itr);
      while(itr2!=l&&*itr2<0){
        *itr-=*itr2;
        auto tmp=itr2;
        itr2=next(tmp);
        ans.erase(itr2);
      }
      ++(*itr);
      if(next(itr)!=l){
        itr=next(itr);
        ans.insert(itr,*itr);
      }
      continue;
    }
    auto itr2=next(itr);
    ll tmp;
    if(*itr2<0){
      tmp=*itr;
    }else{
      tmp=abs(*itr-*itr2);
    }
    if(tmp!=0){
      ans.insert(itr,tmp);
    }else{
      if(itr==ans.begin()||*prev(itr)>0){
        ans.insert(itr,-1);
      }else{
        --(*prev(itr));
      }
    }
    ans.erase(itr);
    itr=itr2;
  }
  if(*itr<0){
    ++(*itr);
    if(*itr==0){
      ans.erase(itr);
    }
  }else{
    ans.erase(itr);
  }
}
void solve(long long N, string a){
  REP(N){
    ans.push_back(a[i]-'0');
  }
  while(ans.size()>1){
    f(ans.end());
  }
  if(ans.empty()||ans.front()<0)cout<<0<<endl;
  else cout<<ans.front()<<endl;

}

int main(){
  cout<<setprecision(1000);
    long long N;
    scanf("%lld",&N);
    string a;
    cin>>a;
    solve(N, std::move(a));
    return 0;
}