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 = "# in the name of god\n\nfrom collections import defaultdict\n\nn, m = list(map(int, input().strip().split()))\n\ng = defaultdict(list)\n\nfor _ in range(m):\n\tu, v = list(map(int, input().strip().split()))\n\tg[u].append(v)\n\tg[v].append(u)\n\nv1 = []\nv2 = []\n\nmark = [-1] * (n + 1)\n\ndef bfs(start):\n    q = []\n    q.append(start)\n    mark[start] = 1\n    v2.append(start)\n\n    while q:\n        v = q.pop(0)\n\n        for u in g[v]:\n            if mark[u] == -1:\n                mark[u] = 1 - mark[v]\n                q.append(u)\n                if mark[u] == 0:\n                    v1.append(u)\n                else:\n                    v2.append(u)\n\n            elif mark[u] == mark[v]:\n                return -1\n    return 0\n\nsw = False\nfor i in range(1, n + 1):\n    if mark[i] == -1 and len(g[i]): \n        res = bfs(i)\n\n    try:\n        if res == -1:\n            sw = True\n            break\n    except:\n        continue\nif sw:\n    print(-1)\nelse:\n    print(len(v1))\n    print(' '.join(str(x) for x in v1))\n    print(len(v2))\n    print(' '.join(str(x) for x in v2))\n\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 14173