Problem:
Score : 400 points

Problem Statement

There are N mountains in a circle, called Mountain 1, Mountain 2, ..., Mountain N in clockwise order. N is an odd number.

Between these mountains, there are N dams, called Dam 1, Dam 2, ..., Dam N. Dam i (1 \leq i \leq N) is located between Mountain i and i+1 (Mountain N+1 is Mountain 1).

When Mountain i (1 \leq i \leq N) receives 2x liters of rain, Dam i-1 and Dam i each accumulates x liters of water (Dam 0 is Dam N).

One day, each of the mountains received a non-negative even number of liters of rain.

As a result, Dam i (1 \leq i \leq N) accumulated a total of A_i liters of water.

Find the amount of rain each of the mountains received. We can prove that the solution is unique under the constraints of this problem.

Constraints

All values in input are integers.

3 \leq N \leq 10^5-1

N is an odd number.

0 \leq A_i \leq 10^9

The situation represented by input can occur when each of the mountains receives a non-negative even number of liters of rain.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print N integers representing the number of liters of rain Mountain 1, Mountain 2, ..., Mountain N received, in this order.

Sample Input 1

3
2 2 4

Sample Output 1

4 0 4

If we assume Mountain 1, 2, and 3 received 4, 0, and 4 liters of rain, respectively, it is consistent with this input, as follows:

Dam 1 should have accumulated \frac{4}{2} + \frac{0}{2} = 2 liters of water.

Dam 2 should have accumulated \frac{0}{2} + \frac{4}{2} = 2 liters of water.

Dam 3 should have accumulated \frac{4}{2} + \frac{4}{2} = 4 liters of water.

Sample Input 2

5
3 8 7 5 5

Sample Output 2

2 4 12 2 8

Sample Input 3

3
1000000000 1000000000 0

Sample Output 3

0 2000000000 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

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

#define rep(i,n) for(int i=0;i<(n);++i)
#define reps(i,n) for(int i=1;i<=(n);++i)
#define repr(i,n) for(int i=(n);i>=0;--i)
#define cinv(n,v) VL v(n);rep(i,n)cin>>v[i]
#define int long long
#define itn int
#define pb push_back
#define eb emplace_back
#define m_p make_pair
#define m_t make_tuple
#define p_p(a,b) pb(m_p(a,b))
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define RSORT(a) sort(all(a),greater<>())
#define UNIQUE(a) a.erase(unique(all(a)),a.end())
#define SZ(a) (signed)(a.size())
#define debg(a) cout<<#a<<" "<<a<<endl;
#define debgp(a) cout<<#a<<" "<<a.fi<<" : "<<a.se<<endl
#define call(a) for(auto i:a)cout<<i<<" ";cout<<endl
#define callp(a) for(auto [k,v]:a)cout<<k<<":"<<v<<endl
#define show(a) for(cont &y:a){for(cont &x:y){cout<<x<<" ";}cout<<endl;}
#define out(a) cout<<(a)<<endl
#define ENDL cout<<endl
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define LINE puts("------------------")
#define fi first
#define se second
#define V vector
#define P pair
#define T tuple
#define PQ priority_queue
#define cont const auto
#define VV(type,name,y,x,a) vector<vector<type>> name = vector<vector<type>>(y,vector<type>(x,a))
typedef long long ll;
typedef double D;
typedef long double LD;
typedef string str;
typedef vector<ll> VL;
typedef pair<ll,ll> PL;
typedef vector<pair<ll,ll>> VP;
typedef tuple<ll,ll,ll> T3;
typedef tuple<ll,ll,ll,ll> T4;
typedef struct{ll to;ll cost;} Graphs;
typedef V<V<Graphs>> AdjList;
const signed INF = (1LL<<30);
const long long MOD = (1e9+7);
const long long LINF = (1LL<<60);
const long long LMAX = LLONG_MAX;
void YN(bool b){cout<<(b?"Yes":"No")<<endl;}

template<class T>inline T gcd(T a,T b){if(b==0)return a; return(gcd(b,a%b));}
template<class T>inline T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T>inline bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
template<class T>inline bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}
template<class T>T center(T a,T b,T c){
  if( (a<=b&&b<=c)||(a>=b&&b>=c) )return b;
  if( (b>=a&&a>=c)||(b<=a&&a<=c) )return a;
  return c;
}
void print(){}
template<class Head,class... Tail>
void print(Head&& head, Tail&&... tail){
  cout<<head<<endl;
  print(forward<Tail>(tail)...);
}
void input(){}
template<class Head,class... Tail>
void input(Head&& head, Tail&&... tail){
  cin>>head;
  input(forward<Tail>(tail)...);
}
template<typename T>
ostream &operator<<(ostream&out,const vector<T>&v){
  if(!v.empty()){
    copy(v.begin(),v.end(),ostream_iterator<T>(out," "));
    out<<"\b\b";
  }
  return out;
}
template<typename T1, typename T2>
ostream &operator<<(ostream&out,const pair<T1,T2>&p){
  out<<p.first<<", "<<p.second;
  return out;
}

pair<ll,ll> func(ll A0,const vector<ll> &a,const ll n){
  ll toL,toR,mid=(n+1)/2;

  vector<ll> AL(mid);
  AL[0]=A0;rep(i,mid-1)AL[i+1]=a[i];
  
  vector<ll> AR(mid);
  ll idx=n-2;
  for(int i=1;i<mid;++i){AR[i]=a[idx];idx--;}
  AR[0]=a[n-1]-A0;
  //print(AL,AR);
  //cout<<"toL ";
  toL=AL[0]*2;
  for(int i=1;i<mid;++i){
    AL[i]-=toL/2;
    //cout<<toL<<" ";
    toL=AL[i]*2;
  }
  //print(toL);
  //cout<<"toR ";
  toR=AR[0]*2;
  for(int i=1;i<mid;++i){
    AR[i]-=toR/2;
    //cout<<toR<<" ";
    toR=AR[i]*2;
  }
  //print(toR);
  //debg(A0);
  return make_pair(toL,toR);
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<fixed<<setprecision(10);
  
  ll n;
  cin>>n;
  vector<ll> a(n);
  rep(i,n)cin>>a[i];

  ll ok=0,ng=a[n-1]+1;
  while(ng-ok>1){
    ll mid=(ok+ng)/2;
    ll toL,toR;
    tie(toL,toR)=func(mid,a,n);
    //print(m_p(toL,toR));
    if(toL==toR){
      ok=mid;
      break;
    }
    else if(toL>toR){
      ng=mid;
    }
    else{
      ok=mid;
    }
  }
  // okをA[i-1]としてシミュレーション
  vector<ll> ans(n);
  //debg(ok);
  ll x=ok*2;
  ans[0]=x;
  for(int i=0;i<n-1;++i){
    a[i]-=x/2;
    x=a[i]*2;
    ans[i+1]=x;
  }
  print(ans);

  return 0;
}