human
The city where Mocha lives in is called Zhijiang. There are n+1 villages and 2n-1 directed roads in this city. 

There are two kinds of roads:

  * n-1 roads are from village i to village i+1, for all 1≤ i ≤ n-1. 
  * n roads can be described by a sequence a_1,…,a_n. If a_i=0, the i-th of these roads goes from village i to village n+1, otherwise it goes from village n+1 to village i, for all 1≤ i≤ n. 


Mocha plans to go hiking with Taki this weekend. To avoid the trip being boring, they plan to go through every village exactly once. They can start and finish at any villages. Can you help them to draw up a plan? 

Input

Each test contains multiple test cases. 

The first line contains a single integer t (1 ≤ t ≤ 20) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains a single integer n (1 ≤ n ≤ 10^4) — indicates that the number of villages is n+1.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1). If a_i=0, it means that there is a road from village i to village n+1. If a_i=1, it means that there is a road from village n+1 to village i.

It is guaranteed that the sum of n over all test cases does not exceed 10^4.

Output

For each test case, print a line with n+1 integers, where the i-th number is the i-th village they will go through. If the answer doesn't exist, print -1.

If there are multiple correct answers, you can print any one of them.

Example

Input

2
3
0 1 0
3
1 1 0

Output

1 4 2 3 
4 1 2 3 

Note

In the first test case, the city looks like the following graph:


So all possible answers are (1 → 4 → 2 → 3), (1 → 2 → 3 → 4).

In the second test case, the city looks like the following graph:


So all possible answers are (4 → 1 → 2 → 3), (1 → 2 → 3 → 4), (3 → 4 → 1 → 2), (2 → 3 → 4 → 1).

Can you provide a Python program to print a line with n+1 integers, where the i-th number is the i-th village they will go through?
assistant
#Here is Python program to print a line with n+1 integers, where the i-th number is the i-th village they will go through.

import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
# sys.setrecursionlimit(111111) 
INF=999999999999999999999999
alphabets="abcdefghijklmnopqrstuvwxyz"
class SortedList:
    def __init__(self, iterable=[], _load=200):
        """Initialize sorted list instance."""
        values = sorted(iterable)
        self._len = _len = len(values)
        self._load = _load
        self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
        self._list_lens = [len(_list) for _list in _lists]
        self._mins = [_list[0] for _list in _lists]
        self._fen_tree = []
        self._rebuild = True
 
    def _fen_build(self):
        """Build a fenwick tree instance."""
        self._fen_tree[:] = self._list_lens
        _fen_tree = self._fen_tree
        for i in range(len(_fen_tree)):
            if i | i + 1 < len(_fen_tree):
                _fen_tree[i | i + 1] += _fen_tree[i]
        self._rebuild = False
 
    def _fen_update(self, index, value):
        """Update `fen_tree[index] += value`."""
        if not self._rebuild:
            _fen_tree = self._fen_tree
            while index < len(_fen_tree):
                _fen_tree[index] += value
                index |= index + 1
 
    def _fen_query(self, end):
        """Return `sum(_fen_tree[:end])`."""
        if self._rebuild:
            self._fen_build()
 
        _fen_tree = self._fen_tree
        x = 0
        while end:
            x += _fen_tree[end - 1]
       ...