ZOJ 2969 Easy Task

Calculating the derivation of a polynomial is an easy task. Given a function f(x) , we use (f(x))' to denote its derivation. We use x^n to denote xn. To calculate the derivation of a polynomial, you should know 3 rules:  (1) (C)'=0 where C is a constant. 

(2) (Cx^n)'=C*n*x^(n-1) where n>=1 and C is a constant. 

(3) (f1(x)+f2(x))'=(f1(x))'+(f2(x))'. It is easy to prove that the derivation a polynomial is also a polynomial.

Here comes the problem, given a polynomial f(x) with non-negative coefficients, can you write a program to calculate the derivation of it?

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 1000) which is the number of test cases. And it will be followed by T consecutive test cases.

There are exactly 2 lines in each test case. The first line of each test case is a single line containing an integer N (0 <= N <= 100). The second line contains N + 1 non-negative integers, CN, CN-1, ..., C1, C0, ( 0 <= Ci <= 1000), which are the coefficients of f(x). Ci is the coefficient of the term with degree i in f(x). (CN!=0)

Output

For each test case calculate the result polynomial g(x) also in a single line. 

(1) If g(x) = 0 just output integer 0.otherwise 

(2) suppose g(x)= Cmx^m+Cm-1x^(m-1)+...+C0 (Cm!=0),then output the integers Cm,Cm-1,...C0.

(3) There is a single space between two integers but no spaces after the last integer. 

Sample Input

3
0
10
2
3 2 1
3
10 0 1 2      

Sample Output

0
6 2

30 0 1      

#include<map>
#include<ctime>
#include<cmath>    
#include<queue> 
#include<string>
#include<vector>
#include<cstdio>    
#include<cstring>  
#include<iostream>
#include<algorithm>    
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))    
#define rep(i,j,k) for(int i=j;i<=k;i++)    
#define per(i,j,k) for(int i=j;i>=k;i--)    
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])    
#define inone(x) scanf("%d",&x)    
#define intwo(x,y) scanf("%d%d",&x,&y)    
#define inthr(x,y,z) scanf("%lf%lf%lf",&x,&y,&z)    
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int T, n, x;

int main()
{
  for (inone(T); T--;)
  {
    inone(n);
    per(i, n, 0)
    {
      inone(x);
      if (i) printf("%d%s", x * i, i == 1 ? "\n" : " ");
    }
    if (!n) puts("0");
  }
  return 0;
}