human
You are given three integers n, a, b. Determine if there exists a permutation p_1, p_2, …, p_n of integers from 1 to n, such that:

  * There are exactly a integers i with 2 ≤ i ≤ n-1 such that p_{i-1} < p_i > p_{i+1} (in other words, there are exactly a local maximums).

  * There are exactly b integers i with 2 ≤ i ≤ n-1 such that p_{i-1} > p_i < p_{i+1} (in other words, there are exactly b local minimums).



If such permutations exist, find any such permutation.

Input

The first line of the input contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. The description of test cases follows.

The only line of each test case contains three integers n, a and b (2 ≤ n ≤ 10^5, 0 ≤ a,b ≤ n).

The sum of n over all test cases doesn't exceed 10^5.

Output

For each test case, if there is no permutation with the requested properties, output -1.

Otherwise, print the permutation that you are found. If there are several such permutations, you may print any of them.

Example

Input

3
4 1 1
6 1 2
6 4 0

Output

1 3 2 4
4 2 3 1 5 6
-1

Note

In the first test case, one example of such permutations is [1, 3, 2, 4]. In it p_1 < p_2 > p_3, and 2 is the only such index, and p_2> p_3 < p_4, and 3 the only such index.

One can show that there is no such permutation for the third test case.
# SHR ... Can you provide a Python program to find any such permutation?
assistant
#Here is Python program to find any such permutation.
i GANESHA    author: Kunal Verma #

import os
import random
import sys

from itertools import *

from io import BytesIO, IOBase

def lcm(a, b):
    return (a * b) // gcd(a, b)

'''
    mod = 10 ** 9 + 7
    fac = [1]
    for i in range(1, 2 * 10 ** 5 + 1):
        fac.append((fac[-1] * i) % mod)
    fac_in = [pow(fac[-1], mod - 2, mod)]
    for i in range(2 * 10 ** 5, 0, -1):
        fac_in.append((fac_in[-1] * i) % mod)
    fac_in.reverse()
    def comb(a, b):
        if a < b:
            return 0
        return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''

# MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
    global spf, adj, MAXN
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i

    for i in range(2, MAXN):
        if i * i > MAXN:
            break
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):
                if (spf[j] == j):
                    spf[j] = i

def getdistinctFactorization(n):
    global adj, spf, MAXN
    for i in range(1, n + 1):
        index = 1
        x = i
        if (x != 1):
            adj[i].append(spf[x])
        x = x // spf[x]
        while (x != 1):
            if (adj[i][index - 1] != spf[x]):
                adj[i].append(spf[x])
                index += 1
            x = x // spf[x]

def printDivisors(n):
    i = 2
    z = [1, n]
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                z.append(i)
            else:
                z.append(i)
                z.append(n // i)
        i = i + 1
    return z

def create(n, x, f):
    pq = len(bin(n)[2:])
    if f == 0:
        tt = min
    else:
        tt = max
    dp = [[inf] * n for _ in range(pq)]
    dp[0] = x
    for i in range(1, pq):
        for j in range(n - (1 << i) + 1):
            dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
    return dp

def enquiry(l, r, dp, f):
    if l > r:
        return inf if not f else -inf
    if f == 1:
        tt = max
    else:
        tt...