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)

#include <bits/stdc++.h>
using namespace std; typedef long double ld; typedef long long ll;
typedef unsigned long long ull;
#define MP make_pair
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define FORR(x,arr) for(auto& x:arr)
#define VI vector<int>
#define PII pair<int, int>
#define FI first 
#define SE second
#define ALL(x) (x).begin(), (x).end()
const int INF=1<<30; const ll LINF=1LL<<60 ; const ll MOD=1e9+7 ;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//-------------------
ll n;
const ll N = 100000 + 5;
ld b[N];
ld c[N+1]; 

int main(){
	cin >> n;
	FOR(i,0,n-1){ cin >> b[i]; b[i] *= 2;}
	FOR(i,0,n-1){
		c[i] = 1.0/2 * (((i%2)==0)?1.0:-1.0);
	}
	//c[n] = 1.0/2;

	/* 
	FOR(i,0,n-1){
		FOR(j,0,n-1){
			cout << a[i][j];
		}
		cout << endl;
	}
	*/
	ld ans[N];
	FOR(i,0,n-1){
		FOR(j,0,n-1){
			ans[i] += c[(j-i+n)%(n)] * b[j];
		}
	}
	//cout << -1%n;
	FOR(i,0,n-1){
		cout << ll(ans[i]) << ' ';
	}
	return 0;
}