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: Time Limit Exceeded)

#pragma region Macros
#include <bits/stdc++.h>
using namespace std;
// iteration helper
#define FOR(i,l,r) for(int i=(l);i<(r);++i)
#define REP(i,n)  FOR(i,0,n)
#define RFOR(i,l,r) for(int i=(r)-1;i>=(l);--i)
#define RREP(i,n) RFOR(i,0,n)
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define PB push_back
// MyVector
template <typename T>
class MyVector : private std::vector<T, std::allocator<T>> {
public:
    using basetype = std::vector<T, std::allocator<T>>;
	using basetype::vector; // constractor
	using basetype::operator=;
	using basetype::begin; using basetype::end; using basetype::cbegin; using basetype::cend; using basetype::rbegin;	using basetype::rend; using basetype::crbegin; using basetype::crend;
	using basetype::size; using basetype::max_size; using basetype::resize; using basetype::capacity; using basetype::empty; using basetype::reserve; using basetype::shrink_to_fit;

	// []のかわりに at をつかう
	typename MyVector<T>::reference operator[](typename basetype::size_type n){ basetype::at(n); }
	typename MyVector<T>::const_reference operator[](typename basetype::size_type n) const { basetype::at(n); };

	using basetype::at; using basetype::data; using basetype::front; using basetype::back;
	using basetype::assign; using basetype::push_back; using basetype::emplace_back; using basetype::pop_back; using basetype::insert; using basetype::emplace; using basetype::erase; using basetype::swap; using basetype::clear;
};
#ifdef _DEBUG
#define vector MyVector
#else
#define vector std::vector
#endif
// typedefs
#define ll long long
#define vi vector<int>
#define vi2 vector<vi>
// input
#define ci(x) int x;cin>>x
#define cs(x) string x;cin>>x
#define cd(x) double x;cin>>x
#define cvi(x,n) vi x(n);REP(i,n){cin>>x[i];}
// const input
#define cci(x) const int x = [](){int t;cin>>t;return t;}()
#define ccs(x) const string x = [](){string t;cin>>t;return t;}()
#define ccd(x) const double x = [](){double t;cin>>t;return t;}()
// output
#define pr(x) cout<<x<<endl
#define prvec(v) REP(i,v.size()) pr(v[i])
#define ppr(x) cout<<x
#define lf() cout<<endl
// debug methods
// usage: debug(x,y);
#define CHOOSE(a) CHOOSE2 a
#define CHOOSE2(a0,a1,a2,a3,a4,x,...) x
#define debug_1(x1) cout<<#x1<<": "<<x1<<endl
#define debug_2(x1,x2) cout<<#x1<<": "<<x1<<", "#x2<<": "<<x2<<endl
#define debug_3(x1,x2,x3) cout<<#x1<<": "<<x1<<", "#x2<<": "<<x2<<", "#x3<<": "<<x3<<endl
#define debug_4(x1,x2,x3,x4) cout<<#x1<<": "<<x1<<", "#x2<<": "<<x2<<", "#x3<<": "<<x3<<", "#x4<<": "<<x4<<endl
#define debug_5(x1,x2,x3,x4,x5) cout<<#x1<<": "<<x1<<", "#x2<<": "<<x2<<", "#x3<<": "<<x3<<", "#x4<<": "<<x4<<", "#x5<<": "<<x5<<endl
#ifdef _DEBUG
#define debug(...) CHOOSE((__VA_ARGS__,debug_5,debug_4,debug_3,debug_2,debug_1,~))(__VA_ARGS__)
#else
#define debug(...)
#endif
// const number
const double PI  = acos(-1.0);
const double EPS = 1e-10;

int powint(int a, int x){
	int res = 1;
	REP(i,x){
		res = res *a; 
	}
	return res;
}
#pragma endregion
#define int long long
signed main(){
#ifdef _DEBUG
try{
#endif
// main program ---------------------------------------------
	cci(N);
	vi A(N);
	REP(i,N){
		cin>>A[i];
	}
	int u = N/2;
	vi x(N);//2x
	REP(i,N){
		int k;
		//if(i==N-1) k=0;
		//else
		k= i+1;
		REP(j,N){
			bool minus = false;
			REP(l,u){
				if(j-k == 2*l || j-k+N == 2*l){
					minus = true;
					//debug(i,j);
				}
			}
			if(minus) x[i] -= A[j];
			else x[i] += A[j];
		}
	}
	REP(i,N){
		cout<<x[i];
		if(i<N-1) cout<<" ";
	}
	cout<<endl;
// end main program -----------------------------------------
#ifdef _DEBUG
} catch (std::out_of_range& ex){ pr("out of range"); }
#endif
	return 0;
}