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>
#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>
#include <numeric>
#include <iomanip>
#include <cstring> 
using namespace std;
using ll=long long;


#define FOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define LLONG_MAXs 9223372036854775800

#define ALL(a) (a).begin(),(a).end()

#include <iostream>
#include <cmath>
using namespace std;
 

 
bool isPrimeNum( ll x ){ // 素数である場合 true を返す
  if( x <= 1 ){ // 1以下である場合は素数でないことがすぐにわかる
    return false;
  }
  // sqrt( double型 ) は引数の平方根を double型で返すので、int型でキャスト
  int n = (int)sqrt( (double)x );
  for( int i = 2; i <= n; i++ ){
    if( x % i == 0 ){ // 割り切る整数がある場合、即判定終了
      return false;
    }
  }
  return true; // 割り切る整数がない場合、素数である
}

ll myPow(ll x, ll n, ll m){
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}

constexpr ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
constexpr ll lcm(ll a,ll b){return a*b/gcd(a,b);}
constexpr ll abs(ll a,ll b){
if(a>=b)return a-b;
if(a<b)return b-a;
}
constexpr ll min(ll a,ll b){
if(a>=b)return b;
if(a<b)return a;
}

constexpr ll max(ll a,ll b){
if(a>=b)return a;
if(a<b)return b;
}

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

class UnionFind {
public:
	//親の番号を格納する。親だった場合は-(その集合のサイズ)
	vector<int> Parent;
 
	//作るときはParentの値を全て-1にする
	//こうすると全てバラバラになる
	UnionFind(int N) {
		Parent = vector<int>(N, -1);
	}
 
	//Aがどのグループに属しているか調べる
	int root(int A) {
		if (Parent[A] < 0) return A;
		return Parent[A] = root(Parent[A]);
	}
 
	//自分のいるグループの頂点数を調べる
	int size(int A) {
		return -Parent[root(A)];//親をとってきたい]
	}
 
	//AとBをくっ付ける
	bool connect(int A, int B) {
		//AとBを直接つなぐのではなく、root(A)にroot(B)をくっつける
		A = root(A);
		B = root(B);
		if (A == B) {
			//すでにくっついてるからくっ付けない
			return false;
		}
 
		//大きい方(A)に小さいほう(B)をくっ付けたい
		//大小が逆だったらひっくり返しちゃう。
		if (size(A) < size(B)) swap(A, B);
 
		//Aのサイズを更新する
		Parent[A] += Parent[B];
		//Bの親をAに変更する
		Parent[B] = A;
 
		return true;
	}
};




long long fac[510000], finv[510000], inv[510000];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < 510000; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
string replaceAll(string &replacedStr, string from, string to) {
    unsigned int pos = replacedStr.find(from);
    int toLen = to.length();
 
    if (from.empty()) {
        return replacedStr;
    }
 
    while ((pos = replacedStr.find(from, pos)) != std::string::npos) {
        replacedStr.replace(pos, from.length(), to);
        pos += toLen;
    }
    return replacedStr;
}

void yn(bool flag) {
    if(flag){
      cout<<"Yes"<<endl;
    }else{
      cout<<"No"<<endl;
    }
    return;
}
void YN(bool flag) {
    if(flag){
      cout<<"YES"<<endl;
    }else{
      cout<<"NO"<<endl;
    }
    return;
}

struct Edge {
    int to;     // 辺の行き先
    int weight; // 辺の重み
    Edge(int t, int w) : to(t), weight(w) { }
};
//using Graph = vector<vector<Edge>>;
using Graph = vector<vector<int>>;

vector<bool> seen;
vector<ll> step;
void dfs(const Graph &G, int v , ll now) {
    seen[v] = true; // v を訪問済にする
    step[v] = now;
    // v から行ける各頂点 next_v について
    for (auto next_v : G[v]) { 
        if (seen[next_v] && (now+1 >= step[next_v]) ) continue; // next_v が探索済だったらスルー
        dfs(G, next_v, now+1); // 再帰的に探索
    }
}

int main() {
  ll N;
  cin>> N;
  ll A[N];
  ll X[N];
  ll S=0;
  ll sum=0;
  FOR(i,0,N-1){
    cin>>A[i];
    S=S+A[i];
    if(i%2!=0 && i>0)sum=sum+A[i];
  }

  X[0]=S-sum*2;

  FOR(i,1,N-1){
    X[i]=A[i-1]*2-X[i-1];
  }

  FOR(i,0,N-1){
    cout<<X[i]<<" ";
  }

  cout<<endl;
  return 0;

}