Problem:
Score : 600 points

Problem Statement

Consider the following arithmetic progression with n terms:

x, x + d, x + 2d, \ldots, x + (n-1)d

What is the product of all terms in this sequence?
Compute the answer modulo 1\ 000\ 003.

You are given Q queries of this form.
In the i-th query, compute the answer in case x = x_i, d = d_i, n = n_i.

Constraints

1 \leq Q \leq 10^5

0 \leq x_i, d_i \leq 1\ 000\ 002

1 \leq n_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

Q
x_1 d_1 n_1
:
x_Q d_Q n_Q

Output

Print Q lines.

In the i-th line, print the answer for the i-th query.

Sample Input 1

2
7 2 4
12345 67890 2019

Sample Output 1

9009
916936

For the first query, the answer is 7 \times 9 \times 11 \times 13 = 9009.
Don't forget to compute the answer modulo 1\ 000\ 003.

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define INF 1<<31-1
#define LLINF 1<<63-1
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define MOD 1000000007
void swap(int* x,int* y){
  int tmp=*x;*x=*y;*y=tmp;
}

typedef struct{int x;int d;int n;LL ans;}P;

int main(){
  int Q,i,j;
  LL ans=0;
  scanf("%d",&Q);
  P q[Q];
  for(i=0;i<Q;i++)scanf("%d%d%d",&q[i].x,&q[i].d,&q[i].n);
  for(i=0;i<Q;i++){
    q[i].ans=1;
    if((q[i].n/1000003)%2==1)q[i].ans=-1;
    q[i].n%=1000003;
    for(j=0;j<q[i].n;j++){
      q[i].ans*=(q[i].x+q[i].d*j);
      if(q[i].ans>=1000003)q[i].ans%=1000003;
    }
  }
  for(i=0;i<Q;i++)printf("%lld\n",q[i].ans);

  //printf("%d\n",ans);
  return 0;
}