Problem: PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: "There exists such a positive integer n that for each positive integer m number n·m + 1 is a prime number".

Unfortunately, PolandBall is not experienced yet and doesn't know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any n.

Input

The only number in the input is n (1 ≤ n ≤ 1000) — number from the PolandBall's hypothesis. 

Output

Output such m that n·m + 1 is not a prime number. Your answer will be considered correct if you output any suitable m such that 1 ≤ m ≤ 103. It is guaranteed the the answer exists.

Examples

Input

3


Output

1

Input

4


Output

2

Note

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

For the first sample testcase, 3·1 + 1 = 4. We can output 1.

In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, m = 2 is okay since 4·2 + 1 = 9, which is not a prime number.

Solution (in PYTHON3):

Solution:
from sys import stdin
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def lsi():
    x=list(stdin.readline())
    x.pop()
    return x
def si(): return stdin.readline()
############# CODE STARTS HERE #############
def sieve(x):
    a=[True]*(x+1)
    sq=floor(sqrt(x))
    for i in range(2, sq+1):
        if a[i]:
            for j in range(i*i, x+1, i):
                a[j]=False
    return a
p=sieve(1000008)
n=ii()
for i in range(1, 1001):
    if not p[n*i+1]:
        print(i)
        exit()