def gcd(x, y):
  if y == 0:
      return x
  else:
      return gcd(y,x%y)

 #最小公倍数
def lcm(a,b):
  return a*(b//gcd(a,b))

N = int(input())
MOD = 998244353
num = lcm(N-1,2*(N-2))
for i in range(3,N//2+1):
  num = lcm(num,i*(N-i))
num %= MOD
print(num)