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>

using namespace std;

vector<int> edge[100050];
int n;
int dep[100050];

void predo(int pos,int lst)
{
	dep[pos] = dep[lst] + 1;
	for(auto v : edge[pos])
		if(v != lst)
			predo(v,pos);
}

int par[100050];

void seek(int pos,int lst)
{
	dep[pos] = dep[lst] + 1;
	par[pos] = lst;
	for(auto v : edge[pos])
		if(v != lst)
			seek(v,pos);
}

vector<int> chain;
bool fg = 0;

void fuck(int pos,int l1,int l2)
{
	for(auto v : edge[pos])
		if(v != l1 && v != l2)
		{
			if(l2 == -1)
			{
				fg = 1;
				return ;
			}
			fuck(v,pos,-1);
		}
}

void Output(int *a)
{
	for(int i = 1;i <= n; ++ i)
		printf("%d ",a[i]);
	exit(0);
}

int main()
{
	scanf("%d",&n);
	for(int i = 1;i < n; ++ i)
	{
		int u,v; scanf("%d%d",&u,&v);
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	predo(1,0);
	int pos = 0,maxn = 0;
	for(int i = 1;i <= n; ++ i)
		if(dep[i] > maxn)
			maxn = dep[i],pos = i;
	int tmp;
	seek(tmp = pos,0);
	pos = 0,maxn = 0;
	for(int i = 1;i <= n; ++ i)
		if(dep[i] > maxn)
			maxn = dep[i],pos = i;
	while(pos != tmp)
	{
		chain.push_back(pos);
		pos = par[pos];
	}
	chain.push_back(tmp);
	static int ans1[100050],ans2[100050];
	int a1c = 0,a2c = 0;
	for(int i = 0;i < chain.size(); ++ i)
	{
		fuck(chain[i],i == 0 ? 0 : chain[i - 1],i == chain.size() - 1 ? 0 : chain[i + 1]);
		if(fg)
		{
			printf("-1\n");
			return 0;
		}
		int c = edge[chain[i]].size() - 2 + (i == 0 || i == chain.size() - 1);
		int l = a1c + 1,r = a1c + 1 + c;
		for(int j = l + 1;j <= r; ++ j)
			ans1[++ a1c] = j;
		ans1[++ a1c] = l;
	}
	for(int i = chain.size() - 1;i >= 0; -- i)
	{
		int c = edge[chain[i]].size() - 2 + (i == 0 || i == chain.size() - 1);
		int l = a2c + 1,r = a2c + 1 + c;
		for(int j = l + 1;j <= r; ++ j)
			ans2[++ a2c] = j;
		ans2[++ a2c] = l;
	}
	for(int i = 1;i <= n; ++ i)
		if(ans1[i] < ans2[i])
			Output(ans1);
		else if(ans1[i] > ans2[i])
			Output(ans2);
	Output(ans1);
}