Codeforces Round #381 (Div. 2) A. Alyona and copybooks

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 for arubles, 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 copybooks k that n + k is divisible by 4? 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 copybooks k that n + k is divisible by 4.

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 have 4 copybooks which she can split between the subjects equally.

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

In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.

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

题意：你原本有n本笔记本，一次买一本笔记本价格a元，两本价格b元，三本价格c元，现在要你买k本，使得（n+k）%4==0，求最少需要花多少钱。

思路：你可以直接暴力枚举所有情况就能出来，情况也不多。我这里是直接背包出答案。下面给代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<queue>
#include<functional>
typedef long long LL;
using namespace std;
#define maxn 200005
#define ll l,mid,now<<1
#define rr mid+1,r,now<<1|1
#define lson l1,mid,l2,r2,now<<1
#define rson mid+1,r1,l2,r2,now<<1|1
#define pi 3.14159
#define inf 0x3f3f3f3f
const int mod = 1e9 + 7;
int main(){
    int n,a[3];
    scanf("%d%d%d%d",&n,&a[0],&a[1],&a[2]);
    n%=4;
    if(n)
        n=4-n;
    LL dp[15];
    memset(dp,inf,sizeof(dp));
    dp[0]=0;
    for(int i=0;i<3;i++){
        for(int j=0;j<=n+8;j++){
            if(j-(i+1)>=0)
                dp[j]=min(dp[j],dp[j-(i+1)]+a[i]);
        }
    }
    LL ans=min(dp[n],dp[n+4]);
    ans=min(ans,dp[n+8]);
    printf("%lld\n",ans);
}