#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const LL N = 300005;

LL ec = 0,hed[N] = {0},nxt[N << 1],to[N << 1];
void add_edge(LL f,LL t){
	++ ec; to[ec] = t; nxt[ec] = hed[f]; hed[f] = ec;
}

LL fa[N],dis[N] = {0},deg[N],vis[N] = {0};
LL a[N],ac = 0,b[N],n;

void dfs(LL u,LL f){
	fa[u] = f; dis[u] = dis[f] + 1;
	deg[u] = 0;
	for(LL i = hed[u];i;i = nxt[i]){
		LL v = to[i];
		if(v == f) continue;
		deg[u] ++;
		dfs(v,u);
	}
}

LL solve(){
	LL id = 0,ed = 0;
	dfs(1,0); for(LL i = 1;i <= n;i ++) if(dis[i] > dis[id]) id = i;
	dfs(id,0); for(LL i = 1;i <= n;i ++) if(dis[i] > dis[ed]) ed = i;
	deg[ed] = 1;
	while(ed){
		a[++ ac] = ed;
		vis[ed] = 1; ed = fa[ed];
	}
	for(LL i = 1;i <= ac;i ++) b[i] = a[ac + 1 - i];
	for(LL i = 1;i <= n;i ++) if(!vis[i] && deg[i]) return 1;
	for(LL i = 1;i <= ac;i ++) a[i] = deg[a[i]],b[i] = deg[b[i]];
	return 0;
}

LL comp(){
	for(LL i = 1;i <= n;i ++){
		if(a[i] > b[i]) return 1;
		if(a[i] < b[i]) return -1;
	}
	return 0;
}

int main(){
	ios::sync_with_stdio(false);
	LL u,v;
	cin >> n;
	for(LL i = 1;i < n;i ++){
		cin >> u >> v;
		add_edge(u,v);
		add_edge(v,u);
	}
	if(solve()){ cout << -1 << '\n'; return 0; }
	if(comp() == 1) swap(a,b);
//	for(LL i = 1;i <= ac;i ++) cout << a[i] << (i == ac ? '\n' : ' ');
	LL now = 1;
	for(LL i = 1;i <= ac;i ++){
		for(LL j = 1;j < a[i];j ++) cout << now + j << ' ';
		cout << now << ' '; now += a[i];
	} cout << '\n';
	return 0;
}
### Prompt

Please create a solution in cpp to 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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const LL N = 300005;

LL ec = 0,hed[N] = {0},nxt[N << 1],to[N << 1];
void add_edge(LL f,LL t){
	++ ec; to[ec] = t; nxt[ec] = hed[f]; hed[f] = ec;
}

LL fa[N],dis[N] = {0},deg[N],vis[N] = {0};
LL a[N],ac = 0,b[N],n;

void dfs(LL u,LL f){
	fa[u] = f; dis[u] = dis[f] + 1;
	deg[u] = 0;
	for(LL i = hed[u];i;i = nxt[i]){
		LL v = to[i];
		if(v == f) continue;
		deg[u] ++;
		dfs(v,u);
	}
}

LL solve(){
	LL id = 0,ed = 0;
	dfs(1,0); for(LL i = 1;i <= n;i ++) if(dis[i] > dis[id]) id = i;
	dfs(id,0); for(LL i = 1;i <= n;i ++) if(dis[i] > dis[ed]) ed = i;
	deg[ed] = 1;
	while(ed){
		a[++ ac] = ed;
		vis[ed] = 1; ed = fa[ed];
	}
	for(LL i = 1;i <= ac;i ++) b[i] = a[ac + 1 - i];
	for(LL i = 1;i <= n;i ++) if(!vis[i] && deg[i]) return 1;
	for(LL i = 1;i <= ac;i ++) a[i] = deg[a[i]],b[i] = deg[b[i]];
	return 0;
}

LL comp(){
	for(LL i = 1;i <= n;i ++){
		if(a[i] > b[i]) return 1;
		if(a[i] < b[i]) return -1;
	}
	return 0;
}

int main(){
	ios::sync_with_stdio(false);
	LL u,v;
	cin >> n;
	for(LL i = 1;i < n;i ++){
		cin >> u >> v;
		add_edge(u,v);
		add_edge(v,u);
	}
	if(solve()){ cout << -1 << '\n'; return 0; }
	if(comp() == 1) swap(a,b);
//	for(LL i = 1;i <= ac;i ++) cout << a[i] << (i == ac ? '\n' : ' ');
	LL now = 1;
	for(LL i = 1;i <= ac;i ++){
		for(LL j = 1;j < a[i];j ++) cout << now + j << ' ';
		cout << now << ' '; now += a[i];
	} cout << '\n';
	return 0;
}
```