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

const int Maxn = 100005;
int n, ct, cnt, head[Maxn], dep[Maxn], p[Maxn], deg[Maxn], ans[Maxn], ans2[Maxn];
bool vis[Maxn];
struct edg
{
	int nxt, to;
}edge[2 * Maxn];
void add(int x, int y)
{
	edge[++cnt] = (edg){head[x], y};
	head[x] = cnt;
}
void dfs(int u, int fa)
{
	dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = edge[i].nxt)
	{
		int to = edge[i].to;
		if (to != fa)
			dfs(to, u);
	}
}
void dfs2(int u, int fa, int dep)
{
	for (int i = head[u]; i; i = edge[i].nxt)
	{
		int to = edge[i].to;
		if (to != fa)
		{
			dfs2(to, u, dep - 1);
			if (ct)
			{
				p[++ct] = u;
				return ;
			}
		}
	}
	if (!dep) p[++ct] = u;
}
void print(int now[])
{
	for (int i = 1; i <= n; i++)
		printf("%d ", now[i]);
	exit(0);
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	int root = 0, maxi = 0;
	for (int i = 1; i <= n; i++)
		if (dep[i] > maxi) maxi = dep[i], root = i;
	dfs(root, 0);
	for (int i = 1; i <= n; i++)
		if (dep[i] > maxi) maxi = dep[i], root = i; 
	dfs2(root, 0, maxi - 1);
	for (int i = 1; i <= ct; i++)
		vis[p[i]] = true;
	for (int u = 1; u <= n; u++)
	{
		if (vis[u]) continue;
		for (int i = head[u]; i; i = edge[i].nxt)
		{
			int to = edge[i].to;
			if (!vis[to])
			{
				puts("-1");
				return 0;
			}
			deg[to]++;
		}
	}
	int now = 0;
	for (int i = 1; i <= ct; i++)
	{
		for (int j = now + 1; j <= now + deg[p[i]]; j++)
			ans[j] = j + 1;
		now += deg[p[i]] + 1;
		ans[now] = now - deg[p[i]];
	}
	reverse(p + 1, p + 1 + ct);
	now = 0;
	for (int i = 1; i <= ct; i++)
	{
		for (int j = now + 1; j <= now + deg[p[i]]; j++)
			ans2[j] = j + 1;
		now += deg[p[i]] + 1;
		ans2[now] = now - deg[p[i]];
	}
	for (int i = 1; i <= n; i++)
		if (ans[i] > ans2[i])
			print(ans2);
		else if (ans[i] < ans2[i])
			print(ans);
	print(ans); 
	return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
int n, ct, cnt, head[Maxn], dep[Maxn], p[Maxn], deg[Maxn], ans[Maxn], ans2[Maxn];
bool vis[Maxn];
struct edg
{
	int nxt, to;
}edge[2 * Maxn];
void add(int x, int y)
{
	edge[++cnt] = (edg){head[x], y};
	head[x] = cnt;
}
void dfs(int u, int fa)
{
	dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = edge[i].nxt)
	{
		int to = edge[i].to;
		if (to != fa)
			dfs(to, u);
	}
}
void dfs2(int u, int fa, int dep)
{
	for (int i = head[u]; i; i = edge[i].nxt)
	{
		int to = edge[i].to;
		if (to != fa)
		{
			dfs2(to, u, dep - 1);
			if (ct)
			{
				p[++ct] = u;
				return ;
			}
		}
	}
	if (!dep) p[++ct] = u;
}
void print(int now[])
{
	for (int i = 1; i <= n; i++)
		printf("%d ", now[i]);
	exit(0);
}
int main()
{
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		add(x, y);
		add(y, x);
	}
	dfs(1, 0);
	int root = 0, maxi = 0;
	for (int i = 1; i <= n; i++)
		if (dep[i] > maxi) maxi = dep[i], root = i;
	dfs(root, 0);
	for (int i = 1; i <= n; i++)
		if (dep[i] > maxi) maxi = dep[i], root = i; 
	dfs2(root, 0, maxi - 1);
	for (int i = 1; i <= ct; i++)
		vis[p[i]] = true;
	for (int u = 1; u <= n; u++)
	{
		if (vis[u]) continue;
		for (int i = head[u]; i; i = edge[i].nxt)
		{
			int to = edge[i].to;
			if (!vis[to])
			{
				puts("-1");
				return 0;
			}
			deg[to]++;
		}
	}
	int now = 0;
	for (int i = 1; i <= ct; i++)
	{
		for (int j = now + 1; j <= now + deg[p[i]]; j++)
			ans[j] = j + 1;
		now += deg[p[i]] + 1;
		ans[now] = now - deg[p[i]];
	}
	reverse(p + 1, p + 1 + ct);
	now = 0;
	for (int i = 1; i <= ct; i++)
	{
		for (int j = now + 1; j <= now + deg[p[i]]; j++)
			ans2[j] = j + 1;
		now += deg[p[i]] + 1;
		ans2[now] = now - deg[p[i]];
	}
	for (int i = 1; i <= n; i++)
		if (ans[i] > ans2[i])
			print(ans2);
		else if (ans[i] < ans2[i])
			print(ans);
	print(ans); 
	return 0;
}

```