#include <algorithm>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <memory>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <list>
 
using namespace std;

typedef long long ll;
 
#define sz size()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(c) (c).begin(), (c).end()
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define clr(a, b) memset((a), (b) ,sizeof(a))
#define ctos(d) string(1,d)
#define print(x) cout<<#x<<" = "<<x<<endl;
 
#define MOD 1000000007

vector<ll> p_decom(ll m){
  vector<ll> v;
  while(m % 2 == 0){
    v.push_back(2);
    m /= 2;
  }
  for(ll i = 3; i * i <= m; i += 2){
    while(m % i == 0){
      v.push_back(i);
      m /= i;
    }
  }
  if(m > 1){
    v.push_back(m);
  }
  sort(all(v));
  return v;
}

ll n;
vector<ll> v;
vector<ll> v1;

ll f1(ll a,ll p){
	ll p1 = 1;
	rep(i,0,v1.sz){
		if(((p>>i)&1)==1){
			p1 *= v1[i];
		}
	}
	if(p1==1){
		return a/p1-1;
	}
	if(a/p1==1){
		return p1-1;
	}
	return p1+a/p1-2;
}

ll f(ll p){
	ll p1 = 1;
	rep(i,0,v.sz){
		if(((p>>i)&1)==1){
			p1 *= v[i];
		}
	}
	ll a = n/p1;
	if(a==1){
		return p1-1;
	}
	v1 = p_decom(a);
	long long left = 1;
	long long right = 1<<(v1.sz)-1;

	while(left != right){
		long long d = (right - left) / 3;
		long long l = f1(a,left + d);
		long long r = f1(a,right - d);
		if(d == 0){
			if(l > r)left++;
			else right--;
		}
		else{
			if(l > r)left = left + d;
			else right = right - d;
		}
	}
	return (p1-1)+f1(a,left);
}

int main(){
	cin>>n;
	v = p_decom(n);
	long long left = 1;
	long long right = 1<<(v.sz)-1;

	while(left != right){
		long long d = (right - left) / 3;
		long long l = f(left + d);
		long long r = f(right - d);
		if(d == 0){
			if(l > r)left++;
			else right--;
		}
		else{
			if(l > r)left = left + d;
			else right = right - d;
		}
	}
	cout << f(left) << " " << n-1 << endl;
	return 0;
}