#include <bits/stdc++.h>
using namespace std;

vector<int> ch[100005];
int fr[100005];
int d[100005];
int n;

void dfs(int at)
{
	for (int u: ch[at]) if (u != fr[at])
	{
		d[u] = d[at]+1;
		fr[u] = at;
		dfs(u);
	}
}

vector<int> calc(const vector<int> &diam)
{
	vector<int> ans;
	int nval = 1;
	for (size_t i = 1;i < diam.size()-1;i++)
	{
		int here = 0;
		for (int u: ch[diam[i]]) if (u != diam[i-1] && u != diam[i+1])
		{
			if (ch[u].size() != 1)
			{
				printf("-1\n");
				exit(0);
			}
			here++;
		}
		for (int j = 1;j <= here;j++) ans.push_back(nval+j);
		ans.push_back(nval);
		nval += here;
		nval++;
	}
	return ans;
}

int main()
{
	scanf("%d", &n);
	for (int i = 1;i < n;i++)
	{
		int a, b; scanf("%d%d", &a, &b);
		--a, --b;
		ch[a].push_back(b);
		ch[b].push_back(a);
	}
	for (int i = 0;i < n;i++) d[i] = 10000000, fr[i] = -1;
	d[0] = 0; dfs(0);
	int s = 0; for (int i = 1;i < n;i++) if (d[i] > d[s]) s = i;
	for (int i = 0;i < n;i++) d[i] = 10000000, fr[i] = -1;
	d[s] = 0; dfs(s);
	s = 0; for (int i = 0;i < n;i++) if (d[i] > d[s]) s = i;
	vector<int> diam;
	diam.push_back(-1);
	for (int at = s;at != -1;at = fr[at]) diam.push_back(at);
	diam.push_back(-1);
	vector<int> ans = calc(diam);
	reverse(diam.begin(), diam.end());
	vector<int> alt = calc(diam);
	ans = min(ans, alt);
	for (int u: ans) printf("%d ", u); printf("\n");
	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;

vector<int> ch[100005];
int fr[100005];
int d[100005];
int n;

void dfs(int at)
{
	for (int u: ch[at]) if (u != fr[at])
	{
		d[u] = d[at]+1;
		fr[u] = at;
		dfs(u);
	}
}

vector<int> calc(const vector<int> &diam)
{
	vector<int> ans;
	int nval = 1;
	for (size_t i = 1;i < diam.size()-1;i++)
	{
		int here = 0;
		for (int u: ch[diam[i]]) if (u != diam[i-1] && u != diam[i+1])
		{
			if (ch[u].size() != 1)
			{
				printf("-1\n");
				exit(0);
			}
			here++;
		}
		for (int j = 1;j <= here;j++) ans.push_back(nval+j);
		ans.push_back(nval);
		nval += here;
		nval++;
	}
	return ans;
}

int main()
{
	scanf("%d", &n);
	for (int i = 1;i < n;i++)
	{
		int a, b; scanf("%d%d", &a, &b);
		--a, --b;
		ch[a].push_back(b);
		ch[b].push_back(a);
	}
	for (int i = 0;i < n;i++) d[i] = 10000000, fr[i] = -1;
	d[0] = 0; dfs(0);
	int s = 0; for (int i = 1;i < n;i++) if (d[i] > d[s]) s = i;
	for (int i = 0;i < n;i++) d[i] = 10000000, fr[i] = -1;
	d[s] = 0; dfs(s);
	s = 0; for (int i = 0;i < n;i++) if (d[i] > d[s]) s = i;
	vector<int> diam;
	diam.push_back(-1);
	for (int at = s;at != -1;at = fr[at]) diam.push_back(at);
	diam.push_back(-1);
	vector<int> ans = calc(diam);
	reverse(diam.begin(), diam.end());
	vector<int> alt = calc(diam);
	ans = min(ans, alt);
	for (int u: ans) printf("%d ", u); printf("\n");
	return 0;
}

```