#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize(2)
typedef long long LL;
/*
懂哥猜想:
规定一个正整数是P数这个正整数当且仅当(是质数或者小于2)
任何大于0的偶数可以写成两个P数之和
*/
vector<LL> prime;int pos;
bitset<100000005> isnotP;
bool isPbaoli(LL n){
    for (LL i=2;i*i<=n;++i){if (!(n%i)){return 0;}}return 1;
}
void euler(LL n){
    pos=0;
    prime.push_back(1145141919810LL);
    for (LL i=1;i<=n;++i){
        if (isPbaoli(n-i)&&(!isnotP[i])){
            printf("%lld %lld",i,n-i);exit(0);
        }
        if ((!isnotP[i])&&(i>=2)){
            prime.push_back(i);++pos;
        }
        for (LL j=1;j<=pos&&i*prime[j]<=n;++j){
            isnotP[i*prime[j]]=1;
            if (!(i%prime[j])){
                break;
            }
        }
    }
}
int main(){
    LL n;
    scanf("%lld",&n);
    euler(n);
    return 0;
}