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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
#define pb push_back
#define ub upper_bound
#define lb upper_bound
#define mp make_pair
#define F first
#define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define SORT(c) sort((c).begin(),(c).end())
#define ve vector
#define vi vector<int>
#define vp vector<pair<int,int>>
#define vvi vector<vector<int>>
 
using ll = long long;
const ll INF = LLONG_MAX - 100;
ll mod = 1e9 + 7;
int dx[] = {-1,0,1,0}, dy[] = {0,1,0,-1};
vector<ll> prime;

class fact {
public:
	int fmod = 1e9+7;
	vector<int> fac, inv;
	fact (int n, int Mod = 1e9+7) {
		fmod = Mod;
		fac = vector<int>(n,0);
		inv = vector<int>(n,0);
		fac[0] = 1; for (int i = 1; i < n; i++) fac[i] = fac[i-1] * i % fmod;
		for (int i = 0;i < n;i++) inv[i] = fact::POW(fac[i],fmod-2);
	}
	ll nCr(ll n, ll r) {return (fac[n] * inv[r] % fmod) * inv[n-r] % fmod;}
	ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%fmod;}a = a * a%fmod; b >>= 1;}return c;}
};
 
template <class T = ll> T in() {T x; cin >> x; return (x);}
void DEBUG(vector<int> a) {for(int i=0;i<a.size();i++)cout<<a[i]<<" ";cout<<endl;}
void EMP(int x) {cout<<"!!!"<<x<<"!!!"<<endl;}
ll GCD(ll a, ll b) {ll c; while (b != 0) {c = a % b; a = b; b = c;}return a;}
ll LCM(ll a, ll b) {return a * b / GCD(a, b);}
ll POW(ll a, ll b, bool usemod = true) {ll c = 1LL; while (b > 0) {if (b & 1LL) {if(!usemod)c*=a;else c = a * c%mod;}if (!usemod) a*=a;else a = a * a%mod; b >>= 1LL;}return c;}
void PRI(ll n) {bool a[n + 1LL]; for (int i = 0; i < n + 1LL; i++) {a[i] = 1LL;}for (int i = 2; i < n + 1LL; i++) {if (a[i]) {prime.pb(i); ll b = i; while (b <= n) {a[b] = 0; b += i;}}}}
template <typename T> T chmin(T& a, T b) {if(a>b)a=b;return a;}
template <typename T> T chmax(T& a, T b) {if(a<b)a=b;return b;}

bool solve() {
	mod = 1000003;
	fact f(2*mod+1,mod);
	int q; cin >> q;
	REP (i,q) {
		int x,d,n; cin >> x >> d >> n;
		if (d == mod) {
			cout << POW(x%d,n) << endl;
			break;
		}
		if (x % mod == 0 || n >= mod) {
			cout << 0 << endl;
			break;
		}
		//fac[x/d+(n-1)] / fac[x/d-1] * d^n;
		int xd = x*POW(d,mod-2)%mod; if (xd == 0) xd = mod;
		int tmp = f.fac[xd+n-1] * f.inv[xd-1] % mod * POW(d,n) % mod;
		cout << tmp << endl;
	}
}

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	solve();
}