1000000000000000
 994
 200385
 3842529393411
 8
 409477218238717
 2
 419477218238718
 909383000
 7
 999088000000000
 265
 415000000238718
 780869426483087
 9
 850085504652042
 567000123
 980123123123123
 409477218238719
 385925923480002
 999998169714888
 936302451686999
 114
 999999993700000
 108000000057
 409477218238716
 112
 1
 409477218238718
 999971000299999
 409477318238718
 735412349812385
 113
 999999999999999
 123830583943
 899990298504716
 936302451687000
 995
 990000000000000
 936302451687001
 999986542686123
 850085504652041

18 999999993541753
 12 941
 14 200355
 17 3842529383076
 7 7
 17 409477218238717
 2 2
 18 419466459294818
 16 909381874
 7 7
 18 999087986204952
 11 212
 18 414993991790735
 18 780869407920631
 7 7
 18 850085504652042
 16 566998782
 18 980123123116482
 18 409477218238718
 17 385925923479720
 18 999998150030846
 18 936302448662019
 11 114
 18 999999993541753
 17 107986074062
 17 409477218238710
 10 106
 1 1
 18 409477218238718
 18 999969994441746
 18 409477218238718
 18 735409591249436
 10 113
 18 999999993541753
 17 123830561521
 18 899973747835553
 18 936302448662019
 12 995
 18 989983621692990
 18 936302448662019
 18 999969994441746
 18 850085504650655

#include <bits/stdc++.h>
using namespace std;
long long a[1000005], p;
long long tri(long long x) { return x * x * x; }
int main() {
  long long re = 1, now = 0, ans1;
  while (1) {
    if (now > 1e15) break;
    if (now + tri(re) < tri(re + 1LL))
      now += tri(re), a[++p] = now;
    else
      re++;
  }
  long long m;
  scanf("%lld", &m);
  p = 19, a[p] = 1e18;
  for (int i = 1; i <= p; i++) {
    if (m == a[i]) {
      ans1 = i;
      break;
    } else if (m < a[i]) {
      ans1 = i - 1;
      break;
    }
  }
  if (m == a[ans1])
    printf("%lld %lld", ans1, a[ans1]);
  else {
    printf("%lld ", ans1);
    for (long long i = 1e5; i; i--)
      if (a[ans1 - 1] + tri(i) <= m)
        return printf("%lld", a[ans1 - 1] + tri(i)), 0;
  }
}
