Codeforce 712A Memory and Crow-白红宇

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.

解题思路:

【题意】

有n个数b1, b2, ..., bn

a1, a2, ..., an是通过等式ai = bi - bi + 1 + bi + 2 - bi + 3....(±)bn得到的

现给你a1, a2, ..., an这n个数,问b1, b2, ..., bn是多少

【类型】

公式推导

【分析】

由此可见,数组b中的第i项等于数组a中的第i项与第i+1项之和

特别地,数组b中的第n项等于数组a中的第n项

【时间复杂度&&优化】

O(n)

题目链接→

1 #include 
       
         2 using namespace std; 3 int main() 4 { 5     int n,a,b; 6     while(cin>>n) 7     { 8         for(int i=1;i<=n;i++) 9         {10             cin>>a;11             if(i>1)12                 cout<
        
         <<" ";13             b=a;14         }15         cout<
         <