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. $u \in A$ or $v \in A$ (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 u_{i} and v_{i} (1  ≤  u_{i},  v_{i}  ≤  n), denoting an undirected edge between u_{i} and v_{i}. 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.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'class Graph:\n\n    def __init__(self):\n        self.numVertices = None\n        self.numEdges = None\n        self.neighbors = None\n        self.loadGraph()\n\n    def loadGraph(self):\n        self.numVertices, self.numEdges = list(map(int, input().split()))\n        self.numVertices += 1\n        self.neighbors = tuple(list() for vertex in range(self.numVertices))\n        self.groups = (set(), set())\n\n        for edge in range(self.numEdges):\n            vertex1, vertex2 = list(map(int, input().split()))\n            self.neighbors[vertex1].append(vertex2)\n            self.neighbors[vertex2].append(vertex1)\n\n    def isBipartite(self):\n        return all(self.isBipartiteComponent(vertex) for vertex in range(1, self.numVertices)\n                                                              if self.groupOf(vertex) == None)\n    def isBipartiteComponent(self, vertex):\n        if len(self.neighbors[vertex]) == 0:\n            return True\n\n        self.groups[0].add(vertex)\n        queue = [vertex]\n\n        while queue:\n            vertex = queue.pop()\n            groupOfVertex = self.groupOf(vertex)\n\n            for neighbor in self.neighbors[vertex]:\n                groupOfNeighbor = self.groupOf(neighbor)\n\n                if groupOfNeighbor == groupOfVertex:\n                    return False\n\n                elif groupOfNeighbor == None:\n                    self.groups[not groupOfVertex].add(neighbor)\n                    queue.append(neighbor)\n\n        return True\n    \n    def groupOf(self, vertex):\n        return 0 if vertex in self.groups[0] else 1 if vertex in self.groups[1] else None\n\n\ndef solve():\n    graph = Graph()\n    if graph.isBipartite():\n        for group in graph.groups:\n            print(len(group))\n            print(" ".join(map(str, group)))\n    else:\n        print(-1)\n\ndef __starting_point():\n    solve()\n\n__starting_point()'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 18583