Problem: We have a secret array. You don't know this array and you have to restore it. However, you know some facts about this array:

  * The array consists of n distinct positive (greater than 0) integers. 
  * The array contains two elements x and y (these elements are known for you) such that x < y. 
  * If you sort the array in increasing order (such that a_1 < a_2 < … < a_n), differences between all adjacent (consecutive) elements are equal (i.e. a_2 - a_1 = a_3 - a_2 = … = a_n - a_{n-1}). 



It can be proven that such an array always exists under the constraints given below.

Among all possible arrays that satisfy the given conditions, we ask you to restore one which has the minimum possible maximum element. In other words, you have to minimize max(a_1, a_2, ..., a_n).

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The only line of the test case contains three integers n, x and y (2 ≤ n ≤ 50; 1 ≤ x < y ≤ 50) — the length of the array and two elements that are present in the array, respectively.

Output

For each test case, print the answer: n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of the required array. If there are several answers, you can print any (it also means that the order of elements doesn't matter).

It can be proven that such an array always exists under the given constraints.

Example

Input


5
2 1 49
5 20 50
6 20 50
5 3 8
9 13 22


Output


1 49 
20 40 30 50 10
26 32 20 38 44 50 
8 23 18 13 3 
1 10 13 4 19 22 25 16 7 

Solution (in PYTHON3):

Solution:
import sys
import math

# Constants
mod = 998244353

# Program Settings
cases = True


# Functions Settings
dynamic_factorial = False



# Pre-defined function
# Begin

# Testcase handler
def main(test_case = False):
    n = int(input()) if test_case else 1
    for _ in range(n):
        test()

# Binary method of power
def fast_pow(a:int, b:int):
    res = 1
    while b > 0:
        if b & 1:
            res *= a
        a *= a
        b >>= 1
    return res

# Combination of 2
def c2(n):
    return n * (n-1) // 2

# Get the list form of input
def get():
    return list(map(int, input().split()))

# Number of 1s in binary form
def bits(n: int):
    return list(bin(n)).count('1')

# For interactive problems
def flush():
    sys.stdout.flush()

# Print array and split it with space
def parr(arr):
    print(*arr, sep=' ')

# GCD (int a, int b)
# Return: value of GCD(a, b)
def gcd(a, b):
    while b:
        if b % a == 0:
            break
        tmp = a
        a = b % a
        b = tmp
    return a

# Extended GCD (int a, int b)
# Return: array of [GCD(a, b), [x0, y0]]
# Where a * x0 + b * y0 = 1
# If GCD(a, b) != 1 then return False
def ext_gcd(a: int, b: int):
    if (b == 0):
        return [a, [1, 0]]
 
    res = ext_gcd(b, a % b)
    g = res[0]
    if g != 1:
        return False
    x1 = res[1][0]
    y1 = res[1][1]
    x = y1
    y = x1 - y1 * (a // b)
 
    return [g, [x, y]]

# Get all different values of array a with their count
def get_set(a):
    res = {}
    for x in a:
        if x in res:
            res[x] += 1
        else:
            res[x] = 1
    return res

# Factorial
fact = {0: 1}
def factorial(n):
    if dynamic_factorial:
        if n not in fact:
            fact[n] = factorial(n-1) * n
        return fact[n]
    else:
        res = 1
        while n > 0:
            res = (res * n) % mod
            n -= 1
        return res

# Sum of array
def sum_of(a):
    res = 0
    for x in a:
        res = (res + x) % mod
    return res

# Prod of array
def prod_of(a):
    res = 1
    for x in a:
        res = (res * x) % mod
    return res

# End

def test():
    n, x, y = get()
    i = 0
    j = 0
    best_i = 0
    best_j = 0
    best_max = 999999999
    while i < n:
        j = i + 1
        while j < n:
            # print("x = {}, y = {}, i = {}, j = {}, n = {}".format(x, y, i, j, n))
            this_max = solve(n, x, y, i, j)
            # print("THIS IS RESULT = {}".format(this_max))
            if (x - i * ((y-x) // (j-i))) <= 0 or (y-x) % (j-i) != 0:
                pass
            elif this_max < best_max:
                best_i = i
                best_j = j
                best_max = this_max
            j += 1
        i += 1
    
    parr(printarr(best_max, n, (y-x)//(best_j - best_i)))

# Problem functions
def solve(n, x, y, i, j):
    return (n-1-j) * ((y-x)//(j-i))+ y

def printarr(max, n, d):
    a = []
    while len(a) < n:
        a.append(max)
        max -= d
    return a

main(cases)