Codeforces 712A Memory and Crow 水题

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, ..., bn written
in a row. For all i from 1 to n,
values ai are
defined by the crows performing the following procedure:

The crow sets ai initially 0.

The crow then adds bi to ai,
subtracts bi + 1,
adds the bi + 2 number,
and so on until the n'th number. Thus, ai = bi - bi + 1 + bi + 2 - bi + 3....

Memory gives you the values a1, a2, ..., an,
and he now wants you to find the initial numbers b1, b2, ..., bn 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 ai ( - 109 ≤ ai ≤ 109) —
the value of the i'th number.

Output

Print n integers corresponding to the sequence b1, b2, ..., bn.
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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
/* run this program using the console pauser or add your own getch, system("pause") or input loop */

ll n;
ll a[100005];
ll re[100005];
int main(int argc, char** argv)
{
scanf("%I64d",&n);
for(int i=1;i<=n;i++)
scanf("%I64d",&a[i]);
re
= a
;
for(int i =1;i<n;i++)
re[i] = a[i]+a[i+1];
for(int i=1;i<=n;i++)
printf("%I64d ",re[i]);
return 0;
}