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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#define scanf __builtin_scanf
#include <algorithm>
using namespace std;

const long mod=1000003;
long fact[1000010];

long modpow(long x, long n)
{
    long res=1;
    for(; n>0; n>>=1) {
        if (n&1) res=res*x%mod;
        x=x*x%mod;
    }
    return res;
}

int main()
{
    fact[0]=1;
    for(int i=1; i<mod; ++i) fact[i]=fact[i-1]*i%mod;

    int Q; scanf("%d", &Q);
    while (Q--) {
        int x, d, n; scanf("%d%d%d", &x, &d, &n);
        if (d==0) {
            printf("%ld\n", modpow(x, n));
            continue;
        }
        auto a=x*modpow(d, mod-2)%mod;
        if (mod-(n-1)<=a or a==0) {
            puts("0");
            continue;
        }
        printf("%ld\n", modpow(d, n)*fact[a+n-1]%mod*modpow(fact[a-1], mod-2)%mod);
    }
}