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:
#from statistics import median
#import collections
#aa = collections.Counter(a) # list to list || .most_common(2)で最大の2個とりだせるお a[0][0]
from fractions import gcd
from itertools import combinations,permutations,accumulate, product # (string,3) 3回
#from collections import deque
from collections import deque,defaultdict,Counter
import decimal
import re
import math
import bisect
#
#
#
# pythonで無理なときは、pypyでやると正解するかも！！
#
#
# my_round_int = lambda x:np.round((x*2 + 1)//2)
# 四捨五入g
#
# インデックス系
# int min_y = max(0, i - 2), max_y = min(h - 1, i + 2);
# int min_x = max(0, j - 2), max_x = min(w - 1, j + 2);
#
#
import sys
sys.setrecursionlimit(10000000)
mod = 10**9 + 7
#mod = 9982443453
#mod = 998244353
from sys import stdin
readline = stdin.readline
def readInts():
  return list(map(int,readline().split()))
def readTuples():
    return tuple(map(int,readline().split()))
def I():
  return int(readline())
def yaku(m):
    ans = []
    i = 1
    while i*i <= m:
        if m % i == 0:
            j = m // i
            ans.append(i)
            if j != i:
                ans.append(j)
        i += 1
    ans = sorted(ans)
    return ans
for _ in range(I()):
    n,x,y = readInts()
    sa = y-x
    if n == 2:
        print(x,y)
        continue
    else:
        lis = yaku(sa)
        # print(lis)
        le = float('inf')
        ans = None
        for a in lis:
            # print(a)
            quo = sa//a
            #print(sa,a,quo,n-2)
            if quo > n-1:
                continue
            else:
                #print(a)
                l = sa // a
                l -= 1
                arr = [x,y]
                for i in range(l):
                    arr.append(x+(i+1)*a)
                nya = n - len(arr)
                small = x-1
                small //= a
                # print(small)
                pien = min(small, nya)
                for j in range(pien):
                    arr.append(x-(j+1)*a)
                nokori = nya - pien
                # print(nokori)
                for k in range(nokori):
                    arr.append(y+(k+1)*a)
                # print(arr)
                if le > max(arr):
                    le = max(arr)
                    ans = arr
        ans = sorted(ans)
        print(*ans)