In the State of Takahashi in AtCoderian Federation, there are N cities, numbered 1, 2, ..., N.
M bidirectional roads connect these cities.
The i-th road connects City A_i and City B_i.
Every road connects two distinct cities.
Also, for any two cities, there is at most one road that directly connects them.
One day, it was decided that the State of Takahashi would be divided into two states, Taka and Hashi.
After the division, each city in Takahashi would belong to either Taka or Hashi.
It is acceptable for all the cities to belong Taka, or for all the cities to belong Hashi.
Here, the following condition should be satisfied:
 - Any two cities in the same state, Taka or Hashi, are directly connected by a road.
Find the minimum possible number of roads whose endpoint cities belong to the same state.
If it is impossible to divide the cities into Taka and Hashi so that the condition is satisfied, print -1.

-----Constraints-----
 - 2 \leq N \leq 700
 - 0 \leq M \leq N(N-1)/2
 - 1 \leq A_i \leq N
 - 1 \leq B_i \leq N
 - A_i \neq B_i
 - If i \neq j, at least one of the following holds: A_i \neq A_j and B_i \neq B_j.
 - If i \neq j, at least one of the following holds: A_i \neq B_j and B_i \neq A_j.

-----Input-----
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
:
A_M B_M

-----Output-----
Print the answer.

-----Sample Input-----
5 5
1 2
1 3
3 4
3 5
4 5

-----Sample Output-----
4

For example, if the cities 1, 2 belong to Taka and the cities 3, 4, 5 belong to Hashi, the condition is satisfied.
Here, the number of roads whose endpoint cities belong to the same state, is 4.

```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 = '"""\nE問題\n繋がっている都市同士は同じ州にあっても違う州にあってもよい→考えにくい\n→「繋がっていない都市は必ず違う州にある」のほうが考えやすい\n→繋がっていない都市にすべて辺を張って二部グラフが構成できるかどうかで判定できそう？\n（必要条件ではあるけど十分条件であるかどうかはわからない）\n→この条件だと同じ州にあるなら必ず任意の二都市間は繋がっているので十分条件も満たしてそう！\n（∵繋がっていない都市に辺を張った時同じ州内に辺を持たないため）\n\n（繋がっていない都市に辺を張った）グラフが連結⇒都市の選び方は高々1通り\n連結でないなら複数の選び方が存在する\n→最小となるような選び方はどのように決めればよいか？\n→それぞれの連結成分ごとに、片方にai都市、もう片方にbi都市振り分けられるとすると\nL = [[a1, b1], [a2, b2], ... , [ak, bk]]\nとできて、(sum(A)+sum(B) == n)\ns = Σ[i = 1 to k](aiかbiのどちらかを選んで足す)\u3000としたときに\nもっともsがn//2に近くなるようにsを選べばよく、これはdpでO(nk)でできる\n"""\nimport sys\ninput = sys.stdin.readline\nsys.setrecursionlimit(10**7)\nimport numpy as np\n\nn, m = map(int, input().split())\nG = [{i for i in range(n) if i != j} for j in range(n)]\nfor _ in range(m):\n  a, b = map(int, input().split())\n  a -= 1\n  b -= 1\n  G[a].remove(b)\n  G[b].remove(a)\n\nseen = [-1]*n\nL = []\ndef is_bipartite(v, color):\n  seen[v] = color\n  L[-1][color] += 1\n  for i in G[v]:\n    if seen[i] != -1:\n      if seen[i] == color:\n        return False\n    else:\n      if not is_bipartite(i, color^1):\n        return False\n  return True\n\nfor i in range(n):\n  if seen[i] == -1:\n    L.append([0, 0])\n    if not is_bipartite(i, 0):\n      print(-1)\n      break\nelse:\n  dp = np.zeros((len(L)+1, n+1), dtype=bool)\n  dp[0, 0] = True\n  for i, ab in enumerate(L):\n    a, b = ab\n    dp[i+1, a:] |= dp[i, :n+1-a]\n    dp[i+1, b:] |= dp[i, :n+1-b]\n  temp = n//2\n  dp = dp[-1].tolist()\n  while not dp[temp]:\n    temp += 1\n  r = n - temp\n  print(temp*(temp-1)//2 + r*(r-1)//2)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 6171301