#include <algorithm>
#include <cstring>
#include <numeric>
#include <cstdio>
int lst[100005], to[200005], pre[200005], tot; 
int ans[100005], seq[100005], deg[100005], dis1[100005], dis2[100005], que[100005], n; 
inline void add_edge(int u, int v)
{
	to[tot] = v; 
	pre[tot] = lst[u]; 
	deg[u]++; 
	lst[u] = tot++; 
}
inline void move(int pos, int val)
{
	int orig = seq[pos]; 
	for (int i = pos; i < pos + val; i++)
		seq[i] = seq[i + 1]; 
	seq[pos + val] = orig; 
}
void calc(int st, int en)
{
	int u = st, sum = 0, x = -1; 
	std::iota(seq, seq + n, 0);
	while (true)
	{
		int val = deg[u] - (u != st) - (u != en); 
		move(sum, val); 
		sum += val + 1; 
		if (u == en)
			break; 
		for (int i = lst[u]; ~i; i = pre[i])
		{
			if (dis1[to[i]] + dis2[to[i]] == dis1[st] + dis1[en] && to[i] != x)
			{
				x = u; 
				u = to[i];
				break; 
			}
		}
	}
	for (int i = 0; i < n; i++)
	{
		if (ans[i] < seq[i])
			return; 
		if (ans[i] > seq[i])
			memcpy(ans, seq, sizeof(seq)); 
	}
}
int bfs(int st, int *dis)
{
	memset(dis, 0x3f, n << 2); 
	dis[st] = 0; 
	int he = 0, ta = 0;
	que[ta++] = st; 
	while (he < ta)
	{
		int u = que[he++]; 
		for (int i = lst[u]; ~i; i = pre[i])
		{
			if (dis[to[i]] > dis[u] + 1)
			{
				dis[to[i]] = dis[u] + 1; 
				que[ta++] = to[i]; 
			}
		}
	}
	return std::max_element(dis, dis + n) - dis; 
}
int main()
{
	// freopen("ARC095-F.in", "r", stdin); 
	memset(lst, -1, sizeof(lst));
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int u, v; 
		scanf("%d%d", &u, &v); 
		add_edge(--u, --v); 
		add_edge(v, u); 
	}
	int u = bfs(0, dis1), v = bfs(u, dis1); 
	bfs(v, dis2); 
	for (int i = 0; i < n; i++)
	{
		bool f = dis1[i] + dis2[i] == dis1[v]; 
		for (int j = lst[i]; ~j && !f; j = pre[j])
		{
			if (dis1[to[j]] + dis2[to[j]] == dis1[v])
				f = true; 
		}
		if (!f)
		{
			puts("-1"); 
			return 0; 
		}
	}
	memset(ans, 0x3f, sizeof(ans)); 
	calc(u, v); 
	calc(v, u); 
	for (int i = 0; i < n; i++)
		printf("%d%c", ans[i] + 1, " \n"[i + 1 == n]);
	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 <algorithm>
#include <cstring>
#include <numeric>
#include <cstdio>
int lst[100005], to[200005], pre[200005], tot; 
int ans[100005], seq[100005], deg[100005], dis1[100005], dis2[100005], que[100005], n; 
inline void add_edge(int u, int v)
{
	to[tot] = v; 
	pre[tot] = lst[u]; 
	deg[u]++; 
	lst[u] = tot++; 
}
inline void move(int pos, int val)
{
	int orig = seq[pos]; 
	for (int i = pos; i < pos + val; i++)
		seq[i] = seq[i + 1]; 
	seq[pos + val] = orig; 
}
void calc(int st, int en)
{
	int u = st, sum = 0, x = -1; 
	std::iota(seq, seq + n, 0);
	while (true)
	{
		int val = deg[u] - (u != st) - (u != en); 
		move(sum, val); 
		sum += val + 1; 
		if (u == en)
			break; 
		for (int i = lst[u]; ~i; i = pre[i])
		{
			if (dis1[to[i]] + dis2[to[i]] == dis1[st] + dis1[en] && to[i] != x)
			{
				x = u; 
				u = to[i];
				break; 
			}
		}
	}
	for (int i = 0; i < n; i++)
	{
		if (ans[i] < seq[i])
			return; 
		if (ans[i] > seq[i])
			memcpy(ans, seq, sizeof(seq)); 
	}
}
int bfs(int st, int *dis)
{
	memset(dis, 0x3f, n << 2); 
	dis[st] = 0; 
	int he = 0, ta = 0;
	que[ta++] = st; 
	while (he < ta)
	{
		int u = que[he++]; 
		for (int i = lst[u]; ~i; i = pre[i])
		{
			if (dis[to[i]] > dis[u] + 1)
			{
				dis[to[i]] = dis[u] + 1; 
				que[ta++] = to[i]; 
			}
		}
	}
	return std::max_element(dis, dis + n) - dis; 
}
int main()
{
	// freopen("ARC095-F.in", "r", stdin); 
	memset(lst, -1, sizeof(lst));
	scanf("%d", &n);
	for (int i = 1; i < n; i++)
	{
		int u, v; 
		scanf("%d%d", &u, &v); 
		add_edge(--u, --v); 
		add_edge(v, u); 
	}
	int u = bfs(0, dis1), v = bfs(u, dis1); 
	bfs(v, dis2); 
	for (int i = 0; i < n; i++)
	{
		bool f = dis1[i] + dis2[i] == dis1[v]; 
		for (int j = lst[i]; ~j && !f; j = pre[j])
		{
			if (dis1[to[j]] + dis2[to[j]] == dis1[v])
				f = true; 
		}
		if (!f)
		{
			puts("-1"); 
			return 0; 
		}
	}
	memset(ans, 0x3f, sizeof(ans)); 
	calc(u, v); 
	calc(v, u); 
	for (int i = 0; i < n; i++)
		printf("%d%c", ans[i] + 1, " \n"[i + 1 == n]);
	return 0; 
}

```