#include <bits/stdc++.h>

using namespace std;

const int N = 100005;

int n;
int root[2];
int par[N];
int nchild[N];
int dis[N];
bool inpath[N];
vector<int> G[N];
vector<int> path;

void dfs(int u, int p, int t) {
	if (dis[root[t]] < dis[u]) root[t] = u;
	par[u] = p, nchild[u] = 1;
	for (auto v : G[u]) if (v != p) {
		dis[v] = dis[u] + 1, dfs(v, u, t), nchild[u] += nchild[v];
	}
}

vector<int> solve() {
	int id = 0;
	vector<int> res;
	for (auto u : path) {
		int tmp = ++id;
		for (auto v : G[u]) {
			if (!inpath[v]) res.push_back(++id);
		}
		res.push_back(tmp);
	}
	return res;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 0; i < (n - 1); ++i) {
		int u, v; cin >> u >> v;
		G[u].push_back(v), G[v].push_back(u);
	}
	root[0] = 1;
	for (int i = 0; i < 2; ++i) {
		dis[root[0]] = 0, dfs(root[0], root[0], 0);
	}
	dis[root[0]] = 0, dfs(root[0], root[0], 1);
	int cur = root[1];
	while (cur != root[0]) {
		path.push_back(cur), inpath[cur] = 1, cur = par[cur];
	}
	path.push_back(root[0]), inpath[root[0]] = 1;
	// for (auto u : path) cout << u << ' '; cout << '\n';
	for (auto u : path) {
		for (auto v : G[u]) {
			if (!inpath[v] && nchild[v] > 1) {
				cout << -1; return 0;
			}
		}
	}
	vector<int> vec0 = solve();
	reverse(path.begin(), path.end());
	vector<int> vec1 = solve();
	if (vec0 > vec1) swap(vec0, vec1);
	for (auto i : vec0) cout << i << ' ';
}
### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

const int N = 100005;

int n;
int root[2];
int par[N];
int nchild[N];
int dis[N];
bool inpath[N];
vector<int> G[N];
vector<int> path;

void dfs(int u, int p, int t) {
	if (dis[root[t]] < dis[u]) root[t] = u;
	par[u] = p, nchild[u] = 1;
	for (auto v : G[u]) if (v != p) {
		dis[v] = dis[u] + 1, dfs(v, u, t), nchild[u] += nchild[v];
	}
}

vector<int> solve() {
	int id = 0;
	vector<int> res;
	for (auto u : path) {
		int tmp = ++id;
		for (auto v : G[u]) {
			if (!inpath[v]) res.push_back(++id);
		}
		res.push_back(tmp);
	}
	return res;
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 0; i < (n - 1); ++i) {
		int u, v; cin >> u >> v;
		G[u].push_back(v), G[v].push_back(u);
	}
	root[0] = 1;
	for (int i = 0; i < 2; ++i) {
		dis[root[0]] = 0, dfs(root[0], root[0], 0);
	}
	dis[root[0]] = 0, dfs(root[0], root[0], 1);
	int cur = root[1];
	while (cur != root[0]) {
		path.push_back(cur), inpath[cur] = 1, cur = par[cur];
	}
	path.push_back(root[0]), inpath[root[0]] = 1;
	// for (auto u : path) cout << u << ' '; cout << '\n';
	for (auto u : path) {
		for (auto v : G[u]) {
			if (!inpath[v] && nchild[v] > 1) {
				cout << -1; return 0;
			}
		}
	}
	vector<int> vec0 = solve();
	reverse(path.begin(), path.end());
	vector<int> vec1 = solve();
	if (vec0 > vec1) swap(vec0, vec1);
	for (auto i : vec0) cout << i << ' ';
}
```