#include <bits/stdc++.h>

using namespace std;

const int MAXN = 100005;

vector<int> G[MAXN];

int deg[MAXN];

int n, u, v;

int p[MAXN];

vector<int> dia;

int dis[MAXN], fa[MAXN], use[MAXN], s, t;

void dfs(int u, int p = 0)
{
	dis[u] = dis[p] + 1, fa[u] = p;
	if (dis[u] > dis[t])
		t = u;
	for (auto v : G[u])
		if (v != p)
			dfs(v, u);
}

void init()
{
	s = 0, t = 0;
	dfs(1);
	s = t, t = 0;
	dfs(s);
	int x = t;
	while (x)
		dia.push_back(x), use[x] = 1, x = fa[x];	
}

int main()
{
	ios::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 1; i < n; i++)
	{
		cin >> u >> v;
		G[u].push_back(v), deg[u]++;
		G[v].push_back(u), deg[v]++;
	}
	init();
	for (auto u : dia)
	{
		for (auto v : G[u])
		{
			if (!use[v] && deg[v] > 1)
			{
				cout << -1 << endl;
				exit(0);	
			}
		}
	}
	int sz = dia.size();
	for (int i = 0; i < sz - i - 1; i++)
	{
		if (deg[dia[i]] < deg[dia[sz - i - 1]])
			break;
		else if (deg[dia[i]] > deg[dia[sz - i - 1]])
		{
			reverse(dia.begin(), dia.end());
			break;
		}
	}
	int sum = 0, val = 0;
	for (int i = 0; i < sz; i++)
	{
		int u = dia[i];
		if (sz > 1)
		{
			if (i == 0 || i == sz - 1)
				deg[u] -= 1;
			else
				deg[u] -= 2;
		}
		p[sum + deg[u] + 1] = ++val;
		for (int j = 1; j <= deg[u]; j++)
			p[sum + j] = ++val;
		sum += deg[u] + 1;
	}
	for (int i = 1; i <= n; i++)
		cout << p[i] << " ";
	cout << endl;
	return 0;	
}

### Prompt

In cpp, your task is to solve the following problem:

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 MAXN = 100005;

vector<int> G[MAXN];

int deg[MAXN];

int n, u, v;

int p[MAXN];

vector<int> dia;

int dis[MAXN], fa[MAXN], use[MAXN], s, t;

void dfs(int u, int p = 0)
{
	dis[u] = dis[p] + 1, fa[u] = p;
	if (dis[u] > dis[t])
		t = u;
	for (auto v : G[u])
		if (v != p)
			dfs(v, u);
}

void init()
{
	s = 0, t = 0;
	dfs(1);
	s = t, t = 0;
	dfs(s);
	int x = t;
	while (x)
		dia.push_back(x), use[x] = 1, x = fa[x];	
}

int main()
{
	ios::sync_with_stdio(false); cin.tie(0);
	cin >> n;
	for (int i = 1; i < n; i++)
	{
		cin >> u >> v;
		G[u].push_back(v), deg[u]++;
		G[v].push_back(u), deg[v]++;
	}
	init();
	for (auto u : dia)
	{
		for (auto v : G[u])
		{
			if (!use[v] && deg[v] > 1)
			{
				cout << -1 << endl;
				exit(0);	
			}
		}
	}
	int sz = dia.size();
	for (int i = 0; i < sz - i - 1; i++)
	{
		if (deg[dia[i]] < deg[dia[sz - i - 1]])
			break;
		else if (deg[dia[i]] > deg[dia[sz - i - 1]])
		{
			reverse(dia.begin(), dia.end());
			break;
		}
	}
	int sum = 0, val = 0;
	for (int i = 0; i < sz; i++)
	{
		int u = dia[i];
		if (sz > 1)
		{
			if (i == 0 || i == sz - 1)
				deg[u] -= 1;
			else
				deg[u] -= 2;
		}
		p[sum + deg[u] + 1] = ++val;
		for (int j = 1; j <= deg[u]; j++)
			p[sum + j] = ++val;
		sum += deg[u] + 1;
	}
	for (int i = 1; i <= n; i++)
		cout << p[i] << " ";
	cout << endl;
	return 0;	
}

```