1. 程式人生 > >【CodeForces - 689C】【Mike and Chocolate Thieves 】(二分)

【CodeForces - 689C】【Mike and Chocolate Thieves 】(二分)

題目:

Bad news came to Mike's village, some thieves stole a bunch of chocolates from the local factory! Horrible!

Aside from loving sweet things, thieves from this area are known to be very greedy. So after a thief takes his number of chocolates for himself, the next thief will take exactly k

 times more than the previous one. The value of k (k > 1) is a secret integer known only to them. It is also known that each thief's bag can carry at most n chocolates (if they intend to take more, the deal is cancelled) and that there were exactly four thieves involved.

Sadly, only the thieves know the value of n, but rumours say that the numbers of ways they could have taken the chocolates (for a fixed n, but not fixed k) is m. Two ways are considered different if one of the thieves (they should be numbered in the order they take chocolates) took different number of chocolates in them.

Mike want to track the thieves down, so he wants to know what their bags are and value of n will help him in that. Please find the smallest possible value of n or tell him that the rumors are false and there is no such n.

Input

The single line of input contains the integer m (1 ≤ m ≤ 1015) — the number of ways the thieves might steal the chocolates, as rumours say.

Output

Print the only integer n — the maximum amount of chocolates that thieves' bags can carry. If there are more than one n satisfying the rumors, print the smallest one.

If there is no such n for a false-rumoured m, print  - 1.

Examples

Input

1

Output

8

Input

8

Output

54

Input

10

Output

-1

Note

In the first sample case the smallest n that leads to exactly one way of stealing chocolates is n = 8, whereas the amounts of stealed chocolates are (1, 2, 4, 8) (the number of chocolates stolen by each of the thieves).

In the second sample case the smallest n that leads to exactly 8 ways is n = 54 with the possibilities: (1, 2, 4, 8),  (1, 3, 9, 27),  (2, 4, 8, 16),  (2, 6, 18, 54),  (3, 6, 12, 24),  (4, 8, 16, 32),  (5, 10, 20, 40),  (6, 12, 24, 48).

There is no n leading to exactly 10 ways of stealing chocolates in the third sample case.

解題報告:相死的心都出來了,英文題居然看不懂,啊啊啊啊,簡單的二分題目。題意大概如下,就是給定有n中投巧克力的方案,且第一個偷x,第二個x*k,第三個 x*k*k 第四個x*k*k*k,問包裹的最小容量。

二分求當前n所能裝納的下的拿巧克力的種類數,然後比較判斷,最後輸出結果 最小的揹包容量,或者-1.

ac程式碼:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;

ll check(ll m)
{
	ll res=0;
	for(ll k=2;k*k*k<=m;k++)
		res+=m/(k*k*k);
	return res;
}
int main()
{
	ll m;
	scanf("%I64d",&m);
	ll res=-1;
	ll l,r,num;
	l=1;r=1e18;
	while(l<=r)
	{
		ll mid=(l+r)/2;
		ll num=check(mid);
		if(num==m)
			res=mid;
		if(num>=m)
			r=mid-1;
		else
			l=mid+1;
	}
	printf("%I64d\n",res);
	
}