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

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本文介绍了一种通过已知变化后的数列逆向求解原始数列的方法。具体地，对于给定的一系列数a1, a2, ..., an，其中每个数ai都是通过特定规则从原始序列b1, b2, ..., bn中计算得出。文章详细解释了如何根据这些变化规律逆向还原出初始数列。

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.

题目大意：

ai=b1-b2+b3-b4+.......bn

给你所有的ai，让你求出bi；

思路：

根据题干中的样例来分析：

a1=b1-b2+b3-b4+b5；

a2=b2-b3+b4-b5；

那么a1+a2=b1；

同理a2+a3=b2；依次类推；

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[100005];
int ans[100005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        ans[n]=a[n];
        for(int i=1;i<=n-1;i++)
        {
            ans[i]=a[i]+a[i+1];
        }
        for(int i=1;i<=n;i++)
        {
            printf("%d ",ans[i]);
        }
        printf("\n");
    }
}