ECC Mini Crypto Challenge - Flag Recovery

from Crypto.Util.number import *
from secret import flag
flag=bytes_to_long(flag)
a =getPrime(256)
b =getPrime(256)
p =getPrime(256)
m1=int(str(flag)[:5])-4585
m2=int(str(flag)[5:])
#EllipticCurve([a1, a2, a3, a4, a6]) -- y^2+(a1)xy+(a3)y=x^3+(a2)x^2+(a4)x+(a6)
E = EllipticCurve(GF(p), [a, b])
X=E.lift_x(m1)
Y=7*X
m = E.random_point()
G = E.random_point()
k = getPrime(256)
K = k * G
r = getPrime(256)
c1 = m + r * K
c2 = r * G
w2=m[0]*m2
print(f"p = {p}")
print(f"a = {a}")
print(f"b = {b}")
print(f"k = {k}")
print(f"E = {E}")
print(f'Y = {Y}')
print(f"c1 = {c1}")
print(f"c2 = {c2}")
print(f"w2 = {w2}")
'''
p = 71397796933602469825964946338224836258949974632540581233301840806613437378503
a = 106105288190268015217241182934677375171023341761047638573248022053052499733117
b = 76170541771321874396004434442157725545076211607587599314450304327736999807927
k = 58155941823118858940343657716409231510854647214870891375273032214774400828217
E = Elliptic Curve defined by y^2 = x^3 + 34707491256665545391276236596452538912073367128507057339946181246439062354614*x + 4772744837719404570039488103932889286126236975047018081148463521123562429424 over Finite Field of size 71397796933602469825964946338224836258949974632540581233301840806613437378503
Y = (33237936857741483513705672980652927705102229733798436323453609986072499230366 : 52619411226266177137991318059937693955038910547834999771526408984808553907338 : 1)
c1 = (37414446283406201193977113266234367761786780230360175925999700345196415953455 : 17037724145039910971426670298726906655653040365428438334942732090559637519851 : 1)
c2 = (60560423732267272277570046154733119097475794979191838027420415113112056962844 : 54372226143125971429691267751299496959531971082475860532181772357190222938465 : 1)
w2 = 16315249811700998894876359855091105114973337718373913477026230968747515636405
'''
# 1. Recover 'r' from c2 and the public point 'G'
r = c2 / G
# 2. Recover 'm' from c1, 'r', and 'K'
m = c1 - r * K
# 3. Recover 'm2' from 'w2' and 'm[0]'
m2 = w2 / m[0]
# 4. Reconstruct the flag from 'm1' and 'm2'
flag_recovered = long_to_bytes(int(str(m1 + 4585) + str(m2)))
print(f'Recovered Flag: {flag_recovered}')

Recovered Flag: b'flag{3cc_15_50_34sy!}'