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
#include <bits/stdc++.h>
#define meow(args...) fprintf(stderr, args)
const int N=1e5+1;
int n, from[N], ans[N];
bool vis[N];
std::vector<int> e[N], lp;
int bfs(int src) {
	static int queue[N];
	int *head=queue, *tail=queue;
	memset(vis+1, 0, n);
	from[*tail++=src]=0;
	while(head!=tail) {
		int u=*head++;
		vis[u]=true;
		for(int v: e[u]) if(!vis[v]) from[*tail++=v]=u;
	}
	return tail[-1];
}
int main() {
	assert(scanf("%d", &n)==1);
	for(int i=1; i<n; ++i) {
		int u, v;
		assert(scanf("%d%d", &u, &v)==2);
		e[u].push_back(v);
		e[v].push_back(u);
	}
	for(int i=bfs(bfs(1)); i; i=from[i]) lp.push_back(i);
	memset(vis+1, 0, n);
	for(int u: lp) for(int v: e[u]) vis[v]=true;
	for(int i=1; i<=n; ++i) if(!vis[i]) {
		puts("-1");
		return 0;
	}
	for(int i=0, j=lp.size(); i<--j; ++i) {
		if(e[lp[i]].size()<e[lp[j]].size()) break;
		if(e[lp[i]].size()>e[lp[j]].size()) {
			std::reverse(lp.begin(), lp.end());
			break;
		}
	}
	for(int i=0, j=0, k=1; i<lp.size(); ++i) {
		int siz=i==0||i==lp.size()-1?0:e[lp[i]].size()-2;
		for(int i=1; i<=siz; ++i) ans[j++]=k+i;
		ans[j++]=k;
		k+=siz+1;
	}
	for(int i=0; i<n; ++i) printf("%d%c", ans[i], " \n"[i==n-1]);
	return 0;
}
