Codeforces Round #280 (Div. 2) A

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题目：

Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.

Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.

Input

The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.

Output

Print the maximum possible height of the pyramid in the single line.

Sample test(s)

input
1

output
1

input
25

output
4

Note

Illustration to the second sample:

没啥好说的。模拟一下就可以了，还wa一发不应该..

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <set>
#include <limits>
#include <map>

using namespace std;

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define F(i, n) for(int (i)=0;(i)<(n);++(i))
#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))
#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))
#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))
#define MEM0(addr) memset((addr), 0, sizeof((addr)))

#define PI 3.1415926535897932384626433832795
#define HALF_PI 1.5707963267948966192313216916398

#define MAXN 100000
#define MAXM 10000
#define MOD 1000000007

typedef long long LL;

const double maxdouble = numeric_limits<double>::max();
const double eps = 1e-10;
const int INF = 0x7FFFFFFF;


int main()
{
    int n;
    cin >> n;
    int ans = 0;
    int base = 1;
    while (n >= base)
    {
        ++ans;
        n -= base;
        base += ans+1;
    }
    cout << ans;
    return 0;
}

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