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>
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define eb emplace_back
#define emp emplace
#define ins insert
#define mp make_pair
#define mt make_tuple
#define sz(s) (int)s.size()
#define forp(i,a,b) for( int i=a;i<=b;i++)
#define rep(i,n)    for( int i=0;i<n;i++)
#define ren(i,n)    for( int i=n-1;i>=0;i--)
#define forn(i,a,b) for( int i=a;i>=b;i--)
#define w(t) while(t)
#define sc(a) scanf("%d", &a);
#define on cout<<"\n"
#define os cout <<" "
#define o2(a,b) cout<<a<<" "<<b
#define o(a) cout << a
#define ppr(a) cout << a.first << " " << a.second
#define bitcount __builtin_popcount
#define gcd __gcd
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(), v.rend()
#define mem(n,m) memset(n,m,sizeof(n))
#define pii pair<int,int>
#define tiii tuple<int, int, int>
#define pll pair<long long,long long>
#define sii set<int>
#define sll set<long long>
#define vii vector<int>
#define vll vector<long long>
#define mii map<int,int>
#define mll map<long long, long long>
#define lob lower_bound
#define upb upper_bound
#define ret return
#define present(s,x) (s.find(x) != s.end())
#define cpresent(s,x) (find(all(s),x) != s.end())
#define ford(container, it) for(auto it = container.begin(); it != container.end(); it++)
#define fors(container, it, a, b) for(auto it = a; it != b; it++)
#define boost ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define MOD 1000000007
#define EPSILON 1e-9
#define PI acos(-1)
#define int long long
#define inf 1e15
#define debug(a) cerr<<#a<<": ";for(auto i:a)cerr<<i<<" ";cerr<<'\n';
#define trace(a) cerr<<#a<<": "<<a<<"\n"
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>


typedef long long  ll;
typedef long double  ldo;
typedef double  db ;
using namespace std;
auto start = std::chrono::system_clock::now();
inline void skj()
{
  std::chrono::time_point<std::chrono::system_clock> end;
  end = std::chrono::system_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  std::time_t end_time = std::chrono::system_clock::to_time_t(end);
  cerr<<"Time taken " << elapsed_seconds.count()*1000<<"\n";
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//uniform_int_distribution<int> ud(1, 100); use this for random number generator

    //custom hash for unordered map
struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};
inline int binexp(int a,int b,int m)
{
    if(a == 0) {
        ret 0;
    }
    int res=1;
    a%=m;
    while(b)
    {
        if(b&1)
            res=(res*1ll*a)%m;
        a=(a*1ll*a)%m;
        b>>=1;
    }
    return res;
}
inline int binmul(int a, int b, int m) {
    a %= m;
    int res = 0;
    w(b) {
        if(b & 1) {
            res = (res + a) % m;
        }
        a = (a + a) % m;
        b >>= 1;
    }
    ret res;
}

inline void solve_me_senpai() {
    int n;
    cin >> n;
    vii arr(n);
    rep(i, n) cin >> arr[i];
    int sum = 0;
    rep(i, n) {
        if(i % 2 == 0) {
            sum += arr[i];
        }
        else {
            sum -= arr[i];
        }
    }
    vii res(1, sum / 2);
    rep(i, n - 1) {
        res.pb(arr[i] -  res.back());
    }
    for(int i : res) {
        cout << 2 * i << " ";
    }
}
int32_t main()
{
    boost
    int t = 1;
    //cin >> t;
    int a = 1;
    while(t--) {
        //cout << "Case #" << a  << ": ";
        solve_me_senpai();
        a++;
    }
    //skj();
    return 0;
}