Problem: Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.

Suppose the graph G is given. Subset A of its vertices is called a vertex cover of this graph, if for each edge uv there is at least one endpoint of it in this set, i.e. <image> or <image> (or both).

Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.

They have agreed to give you their graph and you need to find two disjoint subsets of its vertices A and B, such that both A and B are vertex cover or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — the number of vertices and the number of edges in the prize graph, respectively.

Each of the next m lines contains a pair of integers ui and vi (1 ≤ ui, vi ≤ n), denoting an undirected edge between ui and vi. It's guaranteed the graph won't contain any self-loops or multiple edges.

Output

If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).

If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer k denoting the number of vertices in that vertex cover, and the second line contains k integers — the indices of vertices. Note that because of m ≥ 1, vertex cover cannot be empty.

Examples

Input

4 2
1 2
2 3


Output

1
2 
2
1 3 


Input

3 3
1 2
2 3
1 3


Output

-1

Note

In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to Pari and keep vertex number 4 for yourself (or give it someone, if you wish).

In the second sample, there is no way to satisfy both Pari and Arya.

Solution (in PYTHON3):

Solution:
from sys import stdin
from collections import *


class graph:
    # initialize graph
    def __init__(self, gdict=None):
        if gdict is None:
            gdict = defaultdict(list)
        self.gdict, self.edges, self.arr1, self.arr2, self.visit = gdict, [], defaultdict(int), defaultdict(
            int), defaultdict(int)

    # find edges
    def find_edges(self):
        return self.edges

    # Get verticies
    def get_vertices(self):
        return list(self.gdict.keys())

    # add vertix
    def add_vertix(self, node):
        self.gdict[node] = []

    # add edge
    def add_edge(self, node1, node2):
        self.gdict[node1].append(node2)
        self.gdict[node2].append(node1)
        self.edges.append([node1, node2])

    def bfs(self, i):
        queue = deque([[i, 1]])

        self.visit[i], self.arr1[i] = 1, 1

        while queue:
            # dequeue parent vertix
            s, level = queue.popleft()

            # enqueue child vertices
            for i in self.gdict[s]:
                if self.visit[i] == 0:
                    queue.append([i, level + 1])
                    self.visit[i] = 1
                    if level % 2:
                        self.arr2[i] = 1
                    else:
                        self.arr1[i] = 1


n, m = map(int, input().split())
graph1, e1, e2 = graph(), 0, 0
for i in range(m):
    u, v = map(int, input().split())
    graph1.add_edge(u, v)

for i in range(1, n + 1):
    if not graph1.visit[i]:
        graph1.bfs(i)

out1, out2 = list(graph1.arr1.keys()), list(graph1.arr2.keys())
# print(out1, out2)

for i, j in graph1.edges:
    if graph1.arr1[i] and graph1.arr2[j] or graph1.arr1[j] and graph1.arr2[i]:
        continue
    else:
        exit(print(-1))

print(len(out1))
print(*out1)
print(len(out2))
print(*out2)