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 <iostream>
#include <algorithm>

#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>

#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>

#include<memory>


using namespace std;
#define int long long

typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<double,int> pdi;
typedef pair<long double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()

const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const long double EPS = 1e-12;
const ll mod = 1000000000 + 7;
//const ll mod = 998244353;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const long double pi = 3.141592653589793;

#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

//debug
#define DEBUG

#define DUMPOUT cerr

#ifdef DEBUG
#define dump(...) DUMPOUT<<#__VA_ARGS__<<" :["<<__FUNCTION__<<":"<<__LINE__<<"]"<<endl; DUMPOUT<<"    "; dump_func(__VA_ARGS__)
#else
#define dump(...)
#endif

void dump_func() {DUMPOUT << endl;};

template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) == 0) DUMPOUT << " ";
    else DUMPOUT << ", ";
    dump_func(std::move(tail)...);
}

//ostream
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
    os << "["; for (int i = 0; i<vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : ", "); os << "]";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "[";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")"; itr++;  if(itr != map_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

template<typename T> ostream& operator << (ostream& os, set<T>& set_var) {
    os << "[";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr; ++itr; if(itr != set_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

int n;
int a[100000 + 5];
int sum[100000 + 5];

signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  rep(i, n) {
    cin >> a[i];
    sum[i] += sum[i - 1] + a[i];
  }
  vector<int> ans(n);
  int sum = 0;
  sum += a[0];
  rep2(i, 1, n) {
    if (i % 2 == 1) {
      sum -= a[i];
    }
    else {
      sum += a[i];
    }
  }
  ans[0] = sum;
  rep2(i, 1, n) {
    ans[i] = 2 * a[i - 1] - ans[i - 1];
  }
  rep(i, n) {
    if (i == n -1) cout << ans[i] << endl;
    else cout << ans[i] << " ";
  }
}