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.
Using python3 can you solve the prior task?
 Write me a Python program to find any such permutation?
assistant
#Here is Python program to find any such permutation.

#Fast I/O
import sys,os
import math
# To enable the file I/O i the below 2 lines are uncommented.
# read from in.txt if uncommented
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
# will print on Console if file I/O is not activated
#if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
 
# inputs template
from io import BytesIO, IOBase
from heapq import *
from bisect import *
from collections import *

def main():
    for _ in range(mi()):
        n,a,b = MI()
        x = n-2
        ma = x//2
        mb = x - ma 
        if not ((a <= ma and b <= mb ) or (a <= mb and b <= ma)) or (abs(a-b)>1):
            print(-1)
            continue

        t = (n+1)//2
        arr1 = [i for i in range(1,t + 1)]
        arr2 = [i for i in range(n,n-t+(n%2),-1)]

        if a>=b:
            i = 1
            j = 0
            ans = [arr1[0]]
            flag = 1
        else:
            i = 0
            j = 1
            ans = [arr2[0]]
            flag = 0
        while a != 0 or b != 0 :
            if flag:
                ans.append(arr2[j])
                j+=1
                a-=1
            else:
                ans.append(arr1[i])
                i+=1
                b-=1
            flag^=1
        if flag:
            ans += arr1[i:] + arr2[j:][::-1]
        else:
            ans += arr2[j:] + arr1[i:][::-1]

        print(*ans)

 
# Sample Inputs/Output 
 
# region fastio
 
BUFSIZE = 8192
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
...