CF 680D 堆塔

D. Bear and Tower of Cubes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Limak is a little polar bear. He plays by building towers from blocks. Every block is a cube with positive integer length of side. Limak has infinitely many blocks of each side length.

A block with side a has volume a³. A tower consisting of blocks with sides a₁, a₂, ..., a_k has the total volumea₁³ + a₂³ + ... + a_k³.

Limak is going to build a tower. First, he asks you to tell him a positive integer X — the required total volume of the tower. Then, Limak adds new blocks greedily, one by one. Each time he adds the biggest block such that the total volume doesn't exceed X.

Limak asks you to choose X not greater than m. Also, he wants to maximize the number of blocks in the tower at the end (however, he still behaves greedily). Secondarily, he wants to maximize X.

Can you help Limak? Find the maximum number of blocks his tower can have and the maximum X ≤ m that results this number of blocks.

Input

The only line of the input contains one integer m (1 ≤ m ≤ 10¹⁵), meaning that Limak wants you to choose Xbetween 1 and m, inclusive.

Output

Print two integers — the maximum number of blocks in the tower and the maximum required total volume X, resulting in the maximum number of blocks.

Examples

input

48

output

9 42

input

6

output

6 6

Note

In the first sample test, there will be 9 blocks if you choose X = 23 or X = 42. Limak wants to maximize Xsecondarily so you should choose 42.

In more detail, after choosing X = 42 the process of building a tower is:

• Limak takes a block with side 3 because it's the biggest block with volume not greater than 42. The remaining volume is 42 - 27 = 15.
• The second added block has side 2, so the remaining volume is 15 - 8 = 7.
• Finally, Limak adds 7 blocks with side 1, one by one.

So, there are 9 blocks in the tower. The total volume is is 3³ + 2³ + 7·1³ = 27 + 8 + 7 = 42.

 题意：当确定一个数字x后，这个数字可以用如下的方法取得，找到最大的数字l使得l^3<=x，接下来x-=l^3，这样算一步，

再重复下去直到x==0，得到的总的步数就是数字x的权值，现在给你一个数字n求得数字1-n中权值最大的数字，并且在权值

最大的情况下，让数字的值更大；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int  inf =0x7f7f7f7f;
const double pi=acos(-1);
int p2,q2,p1,ep,q1;
const int INF = 0x3f3f3f3f;
const int maxn = 1100;
pair<int,ll> ans;
ll p3(ll x)
{
    return x*x*x;
}
void dfs(ll n,int p,ll z)
{
    if(!n) {ans=max(ans,make_pair(p,z));return;};
    int t=1;
    while(p3(t+1)<=n) t++;
    dfs(n-p3(t),p+1,z+p3(t));
    dfs(p3(t)-p3(t-1)-1,p+1,z+p3(t-1));
}

int main()
{
    ll n;
    while(~scanf("%lld",&n))
    {
        ans.first=ans.second=0;
        dfs(n,0,0);
        printf("%d %lld
",ans.first,ans.second);
    }
    return 0;
}

 分析：搜索,dfs，复杂度貌似logn