一共三层加密，倒着解密即可

from pwn import xor
import libnum

def decode1(a):
    res = b''
    for i in range(0,63,2):
        res += libnum.n2s(a[i])
    for i in range(63,0,-2):
        res += libnum.n2s(a[i])
    return res

def decode2(a):
    e = 65537
    p = 61571454303089397514579603620349373049341652571832994202527081254304368292533
    q = 59664824358038218622178548968528898444289564045465867369823072940589283303949
    n = p*q
    d=libnum.invmod(e,(p-1)*(q-1))
    return libnum.n2s(pow(a,d,n))

def decode3(string):
    return xor(b'leet', string)

a=open("enc.bin","rb").read()
a=decode1(a)
a=decode2(libnum.s2n(a))
a=decode3(a)
print(a)