Codeforces Round #370 (Div. 2) A

Description

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.

题意：告诉你 a_i = b_i - b_i + 1 + b_i + 2 - b_i + 3 ，和b的序列，求a的序列

解法：反过来就好了，会发现规律都是相加

#include<bits/stdc++.h>
using namespace std;
int a[100005];
int n,m;
int b[100005];
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    int cot=1;
    a[n+1]=0;
    for(int i=n;i>=1;i--)
    {
        b[cot++]=a[i+1]+a[i];
    }
    for(int i=cot-1;i>=1;i--)
    {
        cout<<b[i]<<endl;
    }
    return 0;
}