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 PYTHON):

Solution:
import sys
input = lambda : sys.stdin.readline().rstrip()
def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i: # choufuku
                divisors.append(n//i)
    # divisors.sort()
    return divisors


def answer():
    n, x, y = map(int, input().split(" "))

    diff = y-x
    rs = make_divisors(diff)

    first = True
    for r in rs:
        n1 = diff // r
        if n1+1 > n:
            continue

        if y % r == 0:
            n2 = y // r
            n2 -= 1
        else:
            n2 = y // r
        #print("n2", n2, y, r)

        if n <= n2+1:
            min_max = y
            min_max_lst = [y - i*r for i in reversed(range(n))]
            break
        else:
            lst2 = [y - i*r for i in reversed(range(n2+1))]
            n3 = n - (n2+1)
            lst3 = [y + (i+1)*r for i in range(n3)]
            lst = lst2 + lst3

            cur_max = y + n3 * r

            if first:
                min_max = cur_max
                min_max_lst = lst
                first = False
            else:
                if min_max > cur_max:
                    min_max = cur_max
                    min_max_lst = lst

    for elm in min_max_lst:
        print elm,
    print

  

nn = int(input())
for i in range(nn):
    answer()