human
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.

Example

Input

2
7 2 4
12345 67890 2019

Output

9009
916936
# for
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What is a Python program to print Q lines?
assistant
#Here is Python program to print Q lines.
 i, a in enumerate(iterable)
# q, mod = divmod(a, b)
# divmod(x, y) returns the tuple (x//y, x%y)
# manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
import sys
sys.setrecursionlimit(10**7)
from itertools import combinations, permutations # https://docs.python.org/ja/3/library/itertools.html
from math import factorial
def combinations_count(n, r):
    # faster than the following code
    # def combinations_count(n, r):
    #     return factorial(n) // (factorial(n - r) * factorial(r))
    if n - r < r: r = n - r
    if r == 0: return 1
    if r == 1: return n
    numerator = [n - r + k + 1 for k in range(r)]
    denominator = [k + 1 for k in range(r)]
    for p in range(2,r+1):
        pivot = denominator[p - 1]
        if pivot > 1:
            offset = (n - r) % p
            for k in range(p-1,r,p):
                numerator[k - offset] /= pivot
                denominator[k] /= pivot
    result = 1
    for k in range(r):
        if numerator[k] > 1:
            result *= int(numerator[k])
    return result
def combination_with_repetition_count(n, r):
    return combinations_count(n + r - 1, r)
from collections import deque, Counter # https://docs.python.org/ja/3/library/collections.html#collections.deque
from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
from functools import reduce
from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
def gcds(numbers):
    return reduce(gcd, numbers)
def lcm(x, y):
    return (x * y) // gcd(x, y)
def lcms(numbers):
    return reduce(lcm, numbers, 1)

# first create factorial_list!
# fac_list = mod_factorial_list(n)
INF = 10 ** 18
MOD = 10 ** 6 + 3
def modpow(a, n, p = MOD):
    if   n == 0     : return 1
    elif n == 1     : return a % p
    elif n % 2 == 0 : return modpow(a, n//2, p)**2 % p
    else            : return modpow(a, n//2, p)**2...