'''
解题思路:
构造一个素数判断函数isSu(n),搜索范围为 2 -> int(n**0.5+1)
两个数i、j相加搜索,搜索范围取一半 0 -> n//2+1
'''
def isSu(n):
    for i in range(2,int(n**0.5+1)):
        if n%i==0:
            return False
    return True

while 1:
    try:
        pass

        n = int(input())
        #print('n=',n)
        dis = n
        for i in range(n//2+1):
            #print(i,isSu(i))
            j = n-i
            if isSu(i) and isSu(j) and abs(i-j)<dis:
                i_min = i
                j_min = j

        print(i_min)
        print(j_min)    

    except:
        break