CF1255C.League of Leesins

Bob is an avid fan of the video game "League of Leesins", and today he celebrates as the League of Leesins World Championship comes to an end!

The tournament consisted of $n$ ($n \ge 5$) teams around the world. Before the tournament starts, Bob has made a prediction of the rankings of each team, from $1$-st to $n$-th. After the final, he compared the prediction with the actual result and found out that the $i$-th team according to his prediction ended up at the $p_i$-th position ($1 \le p_i \le n$, all $p_i$ are unique). In other words, $p$ is a permutation of $1, 2, \dots, n$.

As Bob's favorite League player is the famous "3ga", he decided to write down every $3$ consecutive elements of the permutation $p$. Formally, Bob created an array $q$ of $n-2$ triples, where $q_i = (p_i, p_{i+1}, p_{i+2})$ for each $1 \le i \le n-2$. Bob was very proud of his array, so he showed it to his friend Alice.

After learning of Bob's array, Alice declared that she could retrieve the permutation $p$ even if Bob rearranges the elements of $q$ and the elements within each triple. Of course, Bob did not believe in such magic, so he did just the same as above to see Alice's respond.

For example, if $n = 5$ and $p = [1, 4, 2, 3, 5]$, then the original array $q$ will be $[(1, 4, 2), (4, 2, 3), (2, 3, 5)]$. Bob can then rearrange the numbers within each triple and the positions of the triples to get $[(4, 3, 2), (2, 3, 5), (4, 1, 2)]$. Note that $[(1, 4, 2), (4, 2, 2), (3, 3, 5)]$ is not a valid rearrangement of $q$, as Bob is not allowed to swap numbers belong to different triples.

As Alice's friend, you know for sure that Alice was just trying to show off, so you decided to save her some face by giving her any permutation $p$ that is consistent with the array $q$ she was given.

鲍勃是热门电子游戏《英雄联盟：李辛斯》的狂热粉丝，今天他迎来了《英雄联盟：李辛斯》世界锦标赛的圆满落幕！

本次锦标赛共有 $n$（$n \ge 5$）支来自世界各地的战队。在比赛开始前，鲍勃对每支战队的最终排名（从第 $1$ 名到第 $n$ 名）做出了预测。赛后，他将自己的预测结果与实际成绩进行比对，发现按他预测排在第 $i$ 位的战队，实际最终排名为第 $p_i$ 位（$1 \le p_i \le n$，且所有 $p_i$ 互不相同）。换言之，$p$ 是 $1, 2, \dots, n$ 的一个排列。

由于鲍勃最喜爱的《英雄联盟》选手是著名的“3ga”，他决定将排列 $p$ 中所有长度为 $3$ 的连续子序列逐一记录下来。形式化地说，鲍勃构造了一个包含 $n-2$ 个三元组的数组 $q$，其中对每个 $1 \le i \le n-2$，有 $q_i = (p_i, p_{i+1}, p_{i+2})$。鲍勃对自己的这个数组非常自豪，于是把它展示给了他的朋友爱丽丝。

得知鲍勃的数组后，爱丽丝宣称：即使鲍勃打乱数组 $q$ 中各三元组的顺序，并且打乱每个三元组内部三个元素的顺序，她依然能还原出原始排列 $p$。当然，鲍勃并不相信这种“魔法”，于是他照此操作一番，想看看爱丽丝如何应对。

例如，若 $n = 5$ 且 $p = [1, 4, 2, 3, 5]$，则原始数组 $q$ 为 $[(1, 4, 2), (4, 2, 3), (2, 3, 5)]$。鲍勃随后可对每个三元组内部元素重新排序，并对三元组整体顺序重新排列，从而得到 $[(4, 3, 2), (2, 3, 5), (4, 1, 2)]$。注意，$[(1, 4, 2), (4, 2, 2), (3, 3, 5)]$ 并非 $q$ 的合法重排，因为鲍勃不允许将属于不同三元组的数字相互交换。

作为爱丽丝的朋友，你确信她只是在炫耀而已，因此你决定帮她挽回颜面——只需给出任意一个与所给数组 $q$ 相容的排列 $p$ 即可。

The first line contains a single integer $n$ ($5 \le n \le 10^5$) — the size of permutation $p$.

The $i$-th of the next $n-2$ lines contains $3$ integers $q_{i, 1}$, $q_{i, 2}$, $q_{i, 3}$ ($1 \le q_{i, j} \le n$) — the elements of the $i$-th triple of the rearranged (shuffled) array $q_i$, in random order. Remember, that the numbers within each triple can be rearranged and also the positions of the triples can be rearranged.

It is guaranteed that there is at least one permutation $p$ that is consistent with the input.

第一行包含一个整数 $n$（$5 \le n \le 10^5$）—— 表示排列 $p$ 的长度。

接下来的 $n-2$ 行中，第 $i$ 行包含三个整数 $q_{i, 1}$、$q_{i, 2}$、$q_{i, 3}$（$1 \le q_{i, j} \le n$）—— 表示重排（打乱顺序）后的数组 $q_i$ 中第 $i$ 个三元组的三个元素，且这三个元素的顺序是随机的。注意：每个三元组内部的数字顺序可以任意排列，且所有三元组之间的相对位置也可以任意排列。

保证至少存在一个排列 $p$ 满足输入条件。

Print $n$ distinct integers $p_1, p_2, \ldots, p_n$ ($1 \le p_i \le n$) such that $p$ is consistent with array $q$.

If there are multiple answers, print any.

输出 $n$ 个互不相同的整数 $p_1, p_2, \ldots, p_n$（其中 $1 \le p_i \le n$），使得排列 $p$ 与数组 $q$ 一致。

若存在多个答案，输出任意一个即可。