Codeforces#381(Div. 2) A.Alyona and copybooks【暴力】

A. Alyona and copybooks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook fora rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.

What is the minimum amount of rubles she should pay to buy such number of copybooksk that n + k is divisible by4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.

Input

The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 10⁹).

Output

Print the minimum amount of rubles she should pay to buy such number of copybooksk that n + k is divisible by4.

Examples

Input

1 1 3 4

Output

3

Input

6 2 1 1

Output

1

Input

4 4 4 4

Output

0

Input

999999999 1000000000 1000000000 1000000000

Output

1000000000

Note

In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have4 copybooks which she can split between the subjects equally.

In the second example Alyuna can buy a pack of 2 copybooks forb = 1 ruble. She will have 8 copybooks in total.

In the third example Alyona can split the copybooks she already has between the4 subject equally, so she doesn't need to buy anything.

In the fourth example Alyona should buy one pack of one copybook.

题目大意：

我们现在已经有了n个物品，商店卖这种物品有三种方式：

①卖一个，价格为a。

②卖两个，价格为b。

③卖三个，价格为c。

现在要求购买任意方案的数量的该物品，但是需要使得购买量+n是4的倍数。问能够达到目的的最小花费。

思路：

直接暴力枚举每种购买方案的数量即可，比赛中为了稳妥，竟然没有思考的直接裸枚举到了100...............23333333333333333333

其实枚举到4就够了。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define ll __int64int main(){    ll n,a,b,c;    while(~scanf("%I64d%I64d%I64d%I64d",&n,&a,&b,&c))    {        if(n%4==0)        {            printf("0\n");            continue;        }        else        {            ll output=2000000000000000000;            for(ll i=0;i<=100;i++)            {                for(ll j=0;j<=100;j++)                {                    for(ll k=0;k<=100;k++)                    {                        if(((i+j*2+k*3)+n)%4==0)                        {                            output=min(output,a*i+b*j+c*k);                        }                    }                }            }            printf("%I64d\n",output);        }    }}