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;
#define MAXN 100000 
#define rint register int
#define gc() getchar()
inline int read(int r=0,int s=0,int c=gc()){for(;c<48||c>57;s=c,c=gc());for(;c>=48&&c<=57;(r*=10)+=c-48,c=gc());return s^'-'?r:-r;}
int *s, f; inline bool cmps(int i, int j){return s[i]<s[j];}
struct Prog
{
	vector<int> e[MAXN+5]; int n, r, _, d[MAXN+5], s[MAXN+5], c[MAXN+5], Ans[MAXN+5];
	void Find(int p, int fa){(d[p]=d[fa]+1)>d[r]?r=p:0; for(auto v:e[p]) v^fa ? Find(v,p),0 : 0;}
	void DFS_s(int p, int fa){s[p] = 1; for(auto &v:e[p])if(v==fa){swap(v,e[p].back()),e[p].pop_back();break;} for(auto v:e[p]) DFS_s(v,p), s[p] += s[v];}
	void DFS_p(int p){sort(e[p].begin(),e[p].end(),cmps); for(auto v:e[p])if(v^e[p].back()&&s[v]>1) puts("-1"), exit(0); c[p] = ++_; for(auto v:e[p])if(v^e[p].back()) DFS_p(v); if(e[p].size()) DFS_p(e[p].back());}
	void Print(int p){for(auto v:e[p])if(v^e[p].back()) Print(v); Ans[++_] = c[p]; if(!e[p].size()) return; Print(e[p].back());}
}T[2];
int main()
{
	T[0].n = read(); for(rint i = 2, a, b; i <= T[0].n; a = read(), b = read(), T[0].e[a].push_back(b), T[0].e[b].push_back(a), i++); T[1] = T[0];
	s = T[0].s, T[0].Find(1,0), T[0].DFS_s(T[0].r,0), T[0].DFS_p(T[0].r), T[0]._ = 0, T[0].Print(T[0].r), s = T[1].s, T[1].Find(T[0].r,0), T[1].DFS_s(T[1].r,0), T[1].DFS_p(T[1].r), T[1]._ = 0, T[1].Print(T[1].r);
	for(rint i = 1; i <= T[0].n; i++) if(T[0].Ans[i]^T[1].Ans[i]) f = T[0].Ans[i]>T[1].Ans[i]; for(rint i = 1; i <= T[f].n; printf("%d ",T[f].Ans[i]), i++); return 0;
}