Asya loves animals very much. Recently, she purchased nn kittens, enumerated them from 11 and nn and then put them into the cage. The cage consists of one row of nn cells, enumerated with integers from 11 to nn from left to right. Adjacent cells had a partially transparent partition wall between them, hence there were n−1n−1 partitions originally. Initially, each cell contained exactly one kitten with some number.
Observing the kittens, Asya noticed, that they are very friendly and often a pair of kittens in neighboring cells wants to play together. So Asya started to remove partitions between neighboring cells. In particular, on the day ii, Asya:
- Noticed, that the kittens xixi and yiyi, located in neighboring cells want to play together.
- Removed the partition between these two cells, efficiently creating a single cell, having all kittens from two original cells.
Since Asya has never putted partitions back, after n−1n−1 days the cage contained a single cell, having all kittens.
For every day, Asya remembers numbers of kittens xixi and yiyi, who wanted to play together, however she doesn't remember how she placed kittens in the cage in the beginning. Please help her and find any possible initial arrangement of the kittens into nn cells.
Input
The first line contains a single integer nn (2≤n≤1500002≤n≤150000) — the number of kittens.
Each of the following n−1n−1 lines contains integers xixi and yiyi (1≤xi,yi≤n1≤xi,yi≤n, xi≠yixi≠yi) — indices of kittens, which got together due to the border removal on the corresponding day.
It's guaranteed, that the kittens xixi and yiyi were in the different cells before this day.
Output
For every cell from 11 to nn print a single integer — the index of the kitten from 11 to nn, who was originally in it.
All printed integers must be distinct.
It's guaranteed, that there is at least one answer possible. In case there are multiple possible answers, print any of them.
Example
input
Copy
5 1 4 2 5 3 1 4 5
output
Copy
3 1 4 2 5
Note
The answer for the example contains one of several possible initial arrangements of the kittens.
The picture below shows how the cells were united for this initial arrangement. Note, that the kittens who wanted to play together on each day were indeed in adjacent cells.
题意：
给一个长度为n的格子，每个格子之间有挡板，现给出n-1对数 (x y)，每给出一对可撤掉一块挡板，合并x，y所在的两个集合，问原序列是什么
解析：
并查集合并集合+DFS递归输出，我们以样例为例子，设G[ ][ ]为一个图，初始化pre[ i ]==i，表示每个节点祖先都是自身
G[1]={ 4 } 表示1和4相连合并，pre[4]=1
G[2]={ 5 } 表示2和5相连合并，pre[5]=2
G[3]={ 1 } 3和1合并，pre[1]=3
G[4]={ 2 } 4和2合并，为什么不是5呢，因为pre[5]=2，也就是说5有一个前驱是2，所以要率先和前驱合并，pre[ ]的作用就在于此
代码如下：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
vector<int>vec[maxn];
int n;
bool vis[maxn];
int pre[maxn];
int find(int x) {//find函数找前驱
if (x == pre[x])
return x;
return pre[x] = find(pre[x]);
}
void init()
{
for (int i = 0; i <= n; i++)
pre[i] = i;
}
void dfs(int x) {//DFS输出路径
cout << x << ' ';
for (int i = 0; i<vec[x].size(); i++) {
dfs(vec[x][i]);
}
}
int main()
{
ios::sync_with_stdio(false);
cin >> n;
int x, y;
init();
for (int i = 1; i<n; i++) {
cin >> x >> y;
int fx = find(x);//找祖先，也就是前驱
int fy = find(y);
pre[fy] = fx;
vec[fx].push_back(fy);
}
int ans = 0;
for (int i = 1; i <= n; i++) {//第一个节点祖先肯定是自身
if (pre[i] == i) {
ans = i;
break;
}
}
dfs(ans);
return 0;
}