Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction <image> as a sum of three distinct positive fractions in form <image>.

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that <image>. Because Chloe can't check Vladik's answer if the numbers are large, he asks you to print numbers not exceeding 109.

If there is no such answer, print -1.

Input

The single line contains single integer n (1 ≤ n ≤ 104).

Output

If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.

If there are multiple answers, print any of them.

Examples

Input

3


Output

2 7 42


Input

7


Output

7 8 56
import bisect
from itertools import accumulate
import os
import sys
import math
from decimal import *
from io import BytesIO, IOBase
 
BUFSIZE = 8192
 
 
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
 
 
def input():
    return sys.stdin.readline().rstrip("\r\n")
 
 
def isPrime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i = i + 6
    return True
 
 
def SieveOfEratosthenes(n):
    prime = []
    primes = [True for i in range(n + 1)]
    p = 2
    while p * p <= n:
 
        if primes[p] == True:
            prime.append(p)
            for i in range(p * p, n + 1, p):
                primes[i] = False
        p += 1
    return prime
 
 
def primefactors(n):
    fac = []
    while n % 2 == 0:
        fac.append(2)
        n = n // 2
    for i in range(3, int(math.sqrt(n)) + 2):
        while n % i == 0:
            fac.append(i)
            n = n // i
    if n > 1:
        fac.append(n)
    return sorted(fac)
 
 
def factors(n):
    fac = set()
    fac.add(1)
    fac.add(n)
    for i in range(2, int(math.sqrt(n)) + 1):
        if n % i == 0:
            fac.add(i)
            fac.add(n // i)
    return list(fac)
 
 
def modInverse(a, m):
    m0 = m
    y = 0
    x = 1
    if m == 1:
        return 0
    while a > 1:
        q = a // m
        t = m
        m = a % m
        a = t
        t = y
        y = x - q * y
        x = t
    if x < 0:
        x = x + m0
    return x
# ------------------------------------------------------code
n=int(input())
if n==1:
    print(-1)
else:
    print(n,n+1,n*(n+1))