p03384 AtCoder Regular Contest 095 - Permutation Tree
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
```python
import sys
from collections import deque


def diameter(n, links):
    q = deque([(0, -1)])
    v = 0
    while q:
        v, p = q.popleft()
        q.extend((u, v) for u in links[v] if u != p)

    q = deque([(v, -1)])
    w = 0
    parents = [-1] * n
    while q:
        w, p = q.popleft()
        parents[w] = p
        q.extend((u, w) for u in links[w] if u != p)
    parents_rev = [-1] * n
    p = w
    while parents[p] != -1:
        parents_rev[parents[p]] = p
        p = parents[p]
    return v, w, parents, parents_rev


def construct(s, links, parents, parents_rev):
    v = s
    result = []
    while v != -1:
        pv, rv = parents[v], parents_rev[v]
        child_count = 0
        for u in links[v]:
            if u == pv or u == rv:
                continue
            if len(links[u]) != 1:
                return False
            child_count += 1
        my_value = len(result) + 1
        result.extend(range(my_value + 1, my_value + child_count + 1))
        result.append(my_value)
        v = parents[v]
    return result


def solve(n, links):
    d1, d2, parents, parents_rev = diameter(n, links)
    result1 = construct(d1, links, parents_rev, parents)
    if result1 is False:
        return [-1]
    result2 = construct(d2, links, parents, parents_rev)
    return min(result1, result2)


n = int(input())
links = [set() for _ in range(n)]
INF = 10 ** 9

for line in sys.stdin:
    v, w = map(int, line.split())
    v -= 1
    w -= 1
    links[v].add(w)
    links[w].add(v)

print(*solve(n, links))

```