using namespace std;

#include <set>
#include <map>
#include <list>
#include <deque>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <cctype>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <utility>
#include <iomanip>
#include <fstream>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>

#define oo (1<<30)
#define f first
#define s second
#define II inline
#define db double
#define ll long long
#define pb push_back
#define mp make_pair
#define Size(V) ((int)(V.size()))
#define all(V) (V).begin() , (V).end()
#define CC(V) memset((V),0,sizeof((V)))
#define CP(A,B) memcpy((A),(B),sizeof((B)))
#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);++(i))
#define REP(i, N) for (int (i)=0;(i)<(int)(N);++(i))
#define FORit(it, x) for (__typeof((x).begin()) it = (x).begin(); it != (x).end(); ++it)

#define IN  "nummst.in"
#define OUT "nummst.out"

typedef vector<int> VI;
typedef pair<int,int> pi;
typedef vector<string> VS;
template<class T> string toString(T n) {ostringstream ost;ost<<n;ost.flush();return ost.str();}

int N,K,A,B;
int P[1<<10];
int C[4000][21];
int rem[50],Val[4000][50];
bool viz[1<<10];
bool V[21][1<<10];

II void scan()
{
	freopen(IN,"r",stdin);
	freopen(OUT,"w",stdout);
	scanf("%d",&N);
}


II void ciur(int N)
{
	P[++P[0]] = 1; //sometimes 1 can be called a prime number
	P[++P[0]] = 2;
	
	for(int i = 3;i <= N;i += 2)
	{
		if(viz[i] == true)
			continue;
		
		P[++P[0]] = i;
	
		for(int j = i + i;j <= N;j += i)
			viz[j] = true;
	}
	
	FOR(i,2,20)
	{
		V[i][1] = true;
		for(int j = P[i];j <= N / P[i];j *= P[i])
			V[i][j] = true;
	}
}

#define Baza 10000

II void mul(int A[],int B)
{
	int i,b = 0;
	for(i = 1;i <= A[0] || b;++i,b /= Baza)
	{
		b = (B * A[i] + b);
		A[i] = b % Baza;
		b -= (b % Baza);
	}	
	A[0] = i - 1;
}

II bool comp(int A[],int B[])
{
	if(A[0] != B[0])
		return (A[0] > B[0]);
	
	for(int i = A[0];i >= 1;--i)
		if(A[i] != B[i])
			return (A[i] > B[i]);
	
	return false;
}

II void solve()
{
	if(!(N % 2) )
		K = 2;
	for(int k = 3;!K && k * k <= N;k += 2)
		if(!(N % k))
			K = k;
	
	if(K == 2)
	{
		printf("%d %d\n",1 * (N / K),1 * (N / K) );
		return;
	}		
	if(K == 3)
	{
		printf("%d %d\n",2 * (N / K),1 * (N / K) );
		return;
	}
	
	C[1][1] = 1;
	C[2][2] = 1;
	Val[1][0] = Val[1][1] = 1;
	Val[2][0] = 1;
	Val[2][1] = 2;
	
	FOR(i,3,K)
	{
		int nr = 0;
		
		FOR(j,1,P[0])
		{
			if(P[j] >= i)
				break;
			
			if(V[j][ C[ i - P[j] ][j] + 1 ] == false)
				continue;
			
			CP(rem,Val[ i - P[j] ]);
			mul(rem,P[j]);
			
			if( comp(rem,Val[i]) )
			{
				CP(Val[i],rem);
				CP(C[i],C[ i - P[j] ]);
				++C[i][j];
			}
			else
				++nr;
			
			if(nr > 5)
				break;
		}
	}
	
	
	FOR(j,1,20)
		if(C[K][j] != 0)
			printf("%d ",P[j] * C[K][j]	* (N / K) );
	//printf("%d %d\n",(A * 2) * (N / K),(B * 3) * (N / K) );
}

int main()
{
	scan();
	ciur(1<<10);
	solve();
	return 0;
}