#include <bits/stdc++.h>
using namespace std;
#define pb push_back
typedef pair<int,int> pi;
typedef long long ll;

const int N = 1000005;

int n,deg[N],mxNode,mxCost,pr[N];
vector<int> g[N];

void dfs(int u , int p,int dist){
	if(dist>mxCost){
		mxCost = dist;
		mxNode = u;
	}
	pr[u] = p;
	for(auto v : g[u]){
		if(v==p)continue;
		dfs(v,u,dist+1);
	}
}
vector<int> ans;
void append(vector<int> cur){
	for (int i = 0; i < cur.size(); ++i)
		ans.pb(cur[i]);
}
int main(){
	cin >> n;
	for(int u,v,i = 0 ; i < n-1 ; ++i){
		scanf("%d%d",&u,&v);
		g[u].pb(v);
		g[v].pb(u);
	}
	mxNode = mxCost = -1;
	dfs(1,-1,0);
	mxCost = -1;
	dfs(mxNode,-1,0);
	vector<int> all;
	while(1){
		all.push_back(mxNode);
		mxNode = pr[mxNode];
		if(mxNode == -1)break;
	}
	int res = 0;
	for (int i = 0; i < all.size(); ++i){
		deg[all[i]] = g[all[i]].size();
		if(i>0)deg[all[i]]--;
		if(i+1<all.size())deg[all[i]]--;
		res += deg[all[i]]+1;
	}
	if(res != n){
		puts("-1");
		return 0;
	}
	vector<int> fn;
	for(int it = 0 ; it < 2 ; ++it){
		reverse(all.begin(),all.end());
		ans.clear();
		int cnt = n;
		for(int i = 0 ; i < all.size();++i){
			int need = deg[all[i]];
			vector<int> cur;
			for(int j = 0 ; j < need;++j)cur.pb(cnt--);
			reverse(cur.begin(),cur.end());
			cur.push_back(cnt--);
			reverse(cur.begin(),cur.end());
			append(cur);
		}
		reverse(ans.begin(),ans.end());
		if(!it)fn = ans;
		if(it)fn = min(fn,ans);
	}
	for (int i = 0; i < fn.size(); ++i){
		if(i)printf(" ");
		printf("%d",fn[i]);
	}
	puts("");
	return 0;
}
### Prompt

Create a solution in Cpp for the following problem:

Takahashi has an ability to generate a tree using a permutation (p_1,p_2,...,p_n) of (1,2,...,n), in the following process:

First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n, perform the following operation:

* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.



Takahashi is trying to make his favorite tree with this ability. His favorite tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite tree by using a proper permutation. If he can do so, find the lexicographically smallest such permutation.

Constraints

* 2 \leq n \leq 10^5
* 1 \leq v_i, w_i \leq n
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1}


Output

If there is no permutation that can generate a tree isomorphic to Takahashi's favorite tree, print `-1`. If it exists, print the lexicographically smallest such permutation, with spaces in between.

Examples

Input

6
1 2
1 3
1 4
1 5
5 6


Output

1 2 4 5 3 6


Input

6
1 2
2 3
3 4
1 5
5 6


Output

1 2 3 4 5 6


Input

15
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
5 10
5 11
6 12
6 13
7 14
7 15


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
typedef pair<int,int> pi;
typedef long long ll;

const int N = 1000005;

int n,deg[N],mxNode,mxCost,pr[N];
vector<int> g[N];

void dfs(int u , int p,int dist){
	if(dist>mxCost){
		mxCost = dist;
		mxNode = u;
	}
	pr[u] = p;
	for(auto v : g[u]){
		if(v==p)continue;
		dfs(v,u,dist+1);
	}
}
vector<int> ans;
void append(vector<int> cur){
	for (int i = 0; i < cur.size(); ++i)
		ans.pb(cur[i]);
}
int main(){
	cin >> n;
	for(int u,v,i = 0 ; i < n-1 ; ++i){
		scanf("%d%d",&u,&v);
		g[u].pb(v);
		g[v].pb(u);
	}
	mxNode = mxCost = -1;
	dfs(1,-1,0);
	mxCost = -1;
	dfs(mxNode,-1,0);
	vector<int> all;
	while(1){
		all.push_back(mxNode);
		mxNode = pr[mxNode];
		if(mxNode == -1)break;
	}
	int res = 0;
	for (int i = 0; i < all.size(); ++i){
		deg[all[i]] = g[all[i]].size();
		if(i>0)deg[all[i]]--;
		if(i+1<all.size())deg[all[i]]--;
		res += deg[all[i]]+1;
	}
	if(res != n){
		puts("-1");
		return 0;
	}
	vector<int> fn;
	for(int it = 0 ; it < 2 ; ++it){
		reverse(all.begin(),all.end());
		ans.clear();
		int cnt = n;
		for(int i = 0 ; i < all.size();++i){
			int need = deg[all[i]];
			vector<int> cur;
			for(int j = 0 ; j < need;++j)cur.pb(cnt--);
			reverse(cur.begin(),cur.end());
			cur.push_back(cnt--);
			reverse(cur.begin(),cur.end());
			append(cur);
		}
		reverse(ans.begin(),ans.end());
		if(!it)fn = ans;
		if(it)fn = min(fn,ans);
	}
	for (int i = 0; i < fn.size(); ++i){
		if(i)printf(" ");
		printf("%d",fn[i]);
	}
	puts("");
	return 0;
}
```