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> a[111111];
int f(int s, int *p, int &u) {
	int m = -1;
	u = s;
	for (auto i : a[s]) {
		if (i == p[s]) continue;
		p[i] = s;
		int v;
		int j = f(i, p, v);
		if (m < j) m = j, u = v;
	}
	return m + 1;
}
int b[111111];
int main()
{
	int n;
	cin >> n;
	for (int i = 1; i < n; i++) {
		int v, w;
		scanf("%d%d", &v, &w);
		a[v].push_back(w);
		a[w].push_back(v);
	}
	int u, v;
	f(1, b, u);
	b[u] = 0;
	f(u, b, v);
	vector<int> p, q;
	int r = 0;
	n = 1;
	for (int i = v; i != u; i = b[i]) {
		int m = n + 1;
		for (auto j : a[i]) {
			if (j == b[i] || j == r)
				continue;
			if (a[j].size() > 1) {
				cout << -1 << endl;
				return 0;
			}
			p.push_back(m++);
		}
		r = i;
		p.push_back(n);
		n = m;
	}
	p.push_back(n);
	b[v] = 0;
	f(v, b, u);
	r = 0, n = 1;
	for (int i = u; i != v; i = b[i]) {
		int m = n + 1;
		for (auto j : a[i]) {
			if (j == b[i] || j == r)
				continue;
			q.push_back(m++);
		}
		r = i;
		q.push_back(n);
		n = m;
	}
	q.push_back(n);
	p = min(p, q);
	for (auto i : p)
		cout << i << ' ';
	cout << endl;
	return 0;
}
