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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define trace1(a) cout << a << 'n';
#define trace2(a, b) cout << a << ' ' << b << 'n';
#define trace3(a, b, c) cout << a << ' ' << b << ' ' << c << 'n';
#define trace4(a, b, c, d) cout << a << ' ' << b << ' ' << c << ' ' << d << 'n';
#define trace5(a, b, c, d, e) cout << a << ' ' << b << ' ' << c << ' ' << d << ' ' << e << 'n';
#define trace6(a, b, c, d, e, f) cout << a << ' ' << b << ' ' << c << ' ' << d << ' ' << e << ' ' << f << 'n';
#define range(n) for(int i = 0; i < n; i++)
#define printArr(harry, tam) range(tam) cout << harry[i] << " n"[i == tam -1];
#define maxn -2000000 //negative value to break code and force you to put right constraint
#define mod 1000000007
#define md(x) (x) % mod;
#define sc scanf
#define pr printf
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define ps push
#define prt(x) cout << (#x) << " is " << (x) << endl
#define EPS 1e-9
#define INF 1000000000
#define INFd 1e9
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<ll> vll;
typedef vector<double> vd;

int main(){
  ll n;
  cin>>n;
  vector<ll> v(n);
  ll c=0;
  vector<ll> ans(n);
  for(ll i=0;i<n;i++)
    cin>>v[i];
  for(ll i=0;i<n-1;i+=2)
  {
      c+=v[i];
      c-=v[i+1];
  }
  c+=v[n-1];
  ans[0]=c;
  for(ll i=1;i<n;i++)
  {
      ans[i]=2*v[i-1]-ans[i-1];
  }
  for(ll i=0;i<n;i++)
    cout<<ans[i]<<" ";


   return 0;
}