【Codeforces Round 370 (Div 2) A】【水题】Memory and Crow

A. Memory and Crow

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n integers b₁, b₂, ..., b_n written in a row. For all i from 1 to n, values a_i are defined by the crows performing the following procedure:

• The crow sets a_i initially 0.
• The crow then adds b_i to a_i, subtracts b_i + 1, adds the b_i + 2 number, and so on until the n'th number. Thus,a_i = b_i - b_i + 1 + b_i + 2 - b_i + 3....

Memory gives you the values a₁, a₂, ..., a_n, and he now wants you to find the initial numbers b₁, b₂, ..., b_n written in the row? Can you do it?

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of integers written in the row.

The next line contains n, the i'th of which is a_i ( - 10⁹ ≤ a_i ≤ 10⁹) — the value of the i'th number.

Output

Print n integers corresponding to the sequence b₁, b₂, ..., b_n. It's guaranteed that the answer is unique and fits in 32-bit integer type.

Examples

input

5
6 -4 8 -2 3

output

2 4 6 1 3

input

5
3 -2 -1 5 6

output

1 -3 4 11 6

Note

In the first sample test, the crows report the numbers 6, - 4, 8, - 2, and 3 when he starts at indices 1, 2, 3, 4 and 5 respectively. It is easy to check that the sequence 2 4 6 1 3 satisfies the reports. For example, 6 = 2 - 4 + 6 - 1 + 3, and  - 4 = 4 - 6 + 1 - 3.

In the second sample test, the sequence 1,  - 3, 4, 11, 6 satisfies the reports. For example, 5 = 11 - 6 and 6 = 6.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
int a[N];
int b[N];
int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; ++i)scanf("%d", &a[i]);
		a[n + 1] = 0;
		for (int i = n; i >= 1; --i)
		{
			b[i] = a[i] + a[i + 1];
		}
		for (int i = 1; i <= n; ++i)printf("%d ", b[i]);
		puts("");
	}
	return 0;
}
/*
【题意】
a[]和b[]都是长度为n（1e5）的数组，数值范围在[-1e9,1e9]
a[i]=b[i]-b[i+1]+b[i+2]-b[i+3]+...

【类型】
水题

【分析】
直接倒推即可

【时间复杂度&&优化】
O(n)

*/