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)

//Patwari26
#include<bits/stdc++.h>
using namespace std;

#define ll          long long
#define pb          push_back
#define endl        '\n'
#define mii         map<ll,ll>
#define pii         pair<ll,ll>
#define vi          vector<ll>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define INF         (1ll<<60)
#define rep(i,a,b)  for(ll i=a;i<=b;i++)
#define rrep(i,a,b)  for(ll i=a;i>=b;i--)
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define time        cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return a * (b / gcd(a, b)); }
const int N=100005;
void solve(){
    ll n;
    cin>>n;
    vi v(n);
    char c;
    rep(i,0,n-1){
        cin>>c;
        v[i]=c-'0';
    }
    // vi temp;
    // rep(i,0,sz(v)-1) cout<<v[i]<<" \n"[i==(sz(v)-1)];
    // while(v.size()!=1){
    //     rep(i,1,sz(v)-1) temp.pb(abs(v[i]-v[i-1]));
    //     v=temp;
    //     temp.clear();
    //     rep(i,0,sz(v)-1) cout<<v[i]<<" \n"[i==(sz(v)-1)];
    // }
    vi temp;
    while(v.size()!=1){
        rep(i,0,sz(v)-1){
            if(v[i]==0) continue;
            else if(temp.size()<2) temp.pb(v[i]);
            else if(temp[sz(temp)-1]!=v[i] || temp[sz(temp)-2]!=v[i]) temp.pb(v[i]);
        }
        vi fk;
        rep(i,1,sz(temp)-1) fk.pb(abs(temp[i]-temp[i-1]));
        v=fk;
        temp.clear();
    }
    cout<<v[0];
}
signed main()
{
    ios
    int TESTS=1;
    //cin>>TESTS;
    while(TESTS--){
        solve();
    }
    time
    return 0;
}