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: Wrong Answer)

//#pragma once
#include <stdio.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <queue>
#include <algorithm>
#include <sstream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cctype>
#include <cstdlib>      // srand,rand


using namespace std;
#define modd 1000000007



typedef pair<long long, long long> pl;


typedef string::const_iterator State;
class ParseError {};




#define ll long long




class UnionFind {
public:
	vector <ll> par;
	vector <ll> siz;

	// Constructor
	UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL){
		for (ll i = 0; i < sz_; ++i) par[i] = i;
	}
	void init(ll sz_) {
		par.resize(sz_);
		siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
		for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
	}

	// Member Function
	// Find
	ll root(ll x) { // 根の検索
		while (par[x] != x) {
			x = par[x] = par[par[x]]; // x の親の親を x の親とする
		}
		return x;
	}

	// Union(Unite, Merge)
	bool merge(ll x, ll y) {
		x = root(x);
		y = root(y);
		if (x == y) return false;

		// merge technique（データ構造をマージするテク．小を大にくっつける）
		if (siz[x] < siz[y]) swap(x, y);
		siz[x] += siz[y];
		par[y] = x;
		return true;
	}



	bool issame(ll x, ll y) { // 連結判定
		return root(x) == root(y);
	}

	ll size(ll x) { // 素集合のサイズ
		return siz[root(x)];
	}


};


class SegmentTree {
private:
	int n;
	vector <ll> node;

public:
	// 元の配列　v をセグメントツリーで表現する。
	SegmentTree(vector<ll> v) {
		//最下段のノード数は元配列のサイズ以上になる最小の2冪->これをnとおく。
		//セグ木全体で必要なノード数は2n-1個である。
		int sz = v.size();
		n = 1; 
		while (n < sz)n *= 2;

	}

};




long long N, M, K, a, b, c, d, e, H, W, L,T;
ll A[300005] = {};
ll B[300005] = {};
ll C[100][3005] = {};
bool f, ff;
string S,SS;
set <long long>sll;
pl bufpl;
vector <long long>vl[100005];
vector <long long>vll;
vector <pl>vpl;
vector <string> vs;
set<ll> llset[50];
multiset<ll> llmset;


struct ST
{
	ll first;
	ll second;
	ll kaisuu;
	

	bool operator<(const ST& another) const
	{
		return first < another.first;//比較
	};

};



/*vector <ST> vst;
ST st[200005];
ST bufst;
bitset<5000> bits;*/


long long modinv(long long aa, long long mm) {
	long long bb = mm, uu = 1, vv = 0;
	while (bb) {
		long long tt = aa / bb;
		aa -= tt * bb; swap(aa, bb);
		uu -= tt * vv; swap(uu, vv);
	}
	uu %= mm;
	if (uu < 0) uu += mm;
	return uu;
}


ll minn, maxx;
double ansa;

ll zettai(ll aa) {

	if (aa < 0) {
		aa *= -1;
	}
	return aa;

}

ll D;


int main() {
	
	cin >>N;

	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}
	
	a = 0;
	b = 0;
	M = 1;

	if ((N / 2) % 2 == 1) {
		for (int i = 0; i < N / 2; i++) {
			a += A[i]*M;
			b += A[N - 1 - i]*M;
			M *= -1;
		}

		for (ll i = 0; i < 100000; i++) {
			if (a + b - 2 * i*10000< A[N / 2]) {
				e = i;
				break;
			}
		}

		for (ll i = (e-1)*100000; i <= e*100000; i++) {
			if (a+b - 2 * i == A[N / 2]) {
				d = i;
				break;
			}
		}

	}
	else {
		for (int i = 0; i < N / 2; i++) {
			M *= -1;
			a += A[i] * M;
			b += A[N - 1 - i] * M;
			
		}

		for (ll i = 0; i < 100000; i++) {
			if (a + b + 2 * i * 10000 > A[N / 2]) {
				e = i;
				break;
			}
		}

		for (ll i = (e - 1) * 100000; i <= e * 100000; i++) {
			if (a + b + 2 * i == A[N / 2]) {
				d = i;
				break;
			}
		}

	}

	cout << 2 * d << " ";

	for (int i = 0; i < N-1; i++) {
		d = A[i] - d;
		cout << 2 * d;

		if (i == N - 1) {
			cout << endl;
		}
		else {
			cout << " ";
		}
	}

	//cout << fixed << setprecision(10) << ansa << endl;
	return 0;



}