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 MAX_N 100000
using namespace std;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
int n, a, b, p, d[MAX_N+5];
int fa[MAX_N+5], dep[MAX_N+5];
vector <int> G[MAX_N+5], seq, rev, ans;
void DFS(int u) {
	for (int v : G[u]) if (v ^ fa[u])
		fa[v] = u, dep[v] = dep[u]+1, DFS(v);
}
int main() {
	read(n);
	for (int i = 1, u, v; i < n; i++)
		read(u), read(v), d[u]++, d[v]++, 
		G[u].push_back(v), G[v].push_back(u);
	DFS(1);
	for (int i = 1; i <= n; i++)
		if (dep[i] > dep[a]) a = i;
	fa[a] = dep[a] = 0, DFS(a);
	for (int i = 1; i <= n; i++)
		if (dep[i] > dep[b]) b = i;
	for (int i = b; i; i = fa[i])
		if (i == a || i == b)
			seq.push_back(d[i]-1);
		else seq.push_back(d[i]-2);
	rev = seq, reverse(rev.begin(), rev.end());
	for (int i = 0; i < (int)seq.size(); i++)
		if (seq[i] < rev[i]) break;
		else if (seq[i] > rev[i])
			{seq = rev; break;}
	for (int c : seq) {
		p++;
		for (int i = 1; i <= c; i++)
			ans.push_back(p+i);
		ans.push_back(p), p += c;
	}
	if ((int)ans.size() < n)
		return puts("-1"), 0;
	for (int x : ans)
		printf("%d ", x);
	return puts(""), 0;
}
