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;
typedef long long ll;
const ll MOD = 1e9 + 7;
const int INF = 1 << 30;
vector<int> G[100000];
int leaf[100000];
bool vis[100000];
int idx = 1;
void dfs(int v, bool head, vector<int> &L) {
	vis[v] = 1;
	if (head) {
		L.push_back(idx);
		for (int i = 2; i <= leaf[v]; i++) L.push_back(idx + i);
		L.push_back(idx + 1);
		idx += leaf[v] + 1;
		for (int to : G[v]) {
			if (G[to].size() > 1) dfs(to, 0, L);
		}
		return;
	}
	int k = G[v].size() - leaf[v];
	if (k == 1) {
		for (int i = 1; i < leaf[v]; i++) L.push_back(idx + i);
		L.push_back(idx);
		L.push_back(idx + leaf[v]);
		idx += leaf[v] + 1;
		return;
	}
	for (int i = 1; i <= leaf[v]; i++) L.push_back(idx + i);
	L.push_back(idx);
	idx += leaf[v] + 1;
	for (int to : G[v]) {
		if (G[to].size() > 1 && !vis[to]) dfs(to, 0, L);
	}
}
int main() {
	int n;
	cin >> n;
	if (n == 2) {
		cout << "1 2" << endl;
		return 0;
	}
	for (int i = 0; i + 1 < n; i++) {
		int u, v;
		cin >> u >> v; u--; v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	vector<int> term;
	for (int i = 0; i < n; i++) {
		if (G[i].size() == 1) continue;
		for (int to : G[i]) {
			if (G[to].size() == 1) leaf[i]++;
		}
		int k = G[i].size() - leaf[i];
		if (k == 1) term.push_back(i);
		else if (k >= 3) {
			cout << -1 << endl;
			return 0;
		}
	}
	if (term.empty()) { //うに
		cout << 1;
		for (int i = 3; i < n; i++) cout << " " << i;
		cout << " 2 " << n << endl;
		return 0;
	}
	vector<int> L1, L2;
	idx = 1;
	memset(vis, 0, sizeof(vis));
	dfs(term[0], 1, L1);
	idx = 1;
	memset(vis, 0, sizeof(vis));
	dfs(term[1], 1, L2);
	vector<int> L = min(L1, L2);
	for (int i = 0; i < L.size(); i++) {
		cout << L[i] << " ";
	}
	cout << endl;
}
