Codeforces Round #478 (Div. 2) B. Mancala

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本文介绍了一款流行于中东地区的棋盘游戏Mancala及其算法解决方案。玩家通过选择并重新分配棋盘上特定坑洞中的石头来获取分数，目标是在一轮行动后获得最高分。文章详细解释了游戏规则，并提供了一个C++实现的例子，用于计算最大可能得分。

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mancala is a game famous in the Middle East. It is played on a board that consists of 14 holes.

Initially, each hole has ai

stones. When a player makes a move, he chooses a hole which contains a positive number of stones. He takes all the stones inside it and then redistributes these stones one by one in the next holes in a counter-clockwise direction.

Note that the counter-clockwise order means if the player takes the stones from hole i

, he will put one stone in the (i+1)-th hole, then in the (i+2)-th, etc. If he puts a stone in the 14

-th hole, the next one will be put in the first hole.

After the move, the player collects all the stones from holes that contain even number of stones. The number of stones collected by player is the score, according to Resli.

Resli is a famous Mancala player. He wants to know the maximum score he can obtain after one move.

Input

The only line contains 14 integers a1,a2,…,a14

(0≤ai≤109

) — the number of stones in each hole.

It is guaranteed that for any i

(1≤i≤14) ai

is either zero or odd, and there is at least one stone in the board.

Output

Output one integer, the maximum possible score after one move.

Examples

Input

Copy

0 1 1 0 0 0 0 0 0 7 0 0 0 0

Output

Copy

Input

Copy

5 1 1 1 1 0 0 0 0 0 0 0 0 0

Output

Copy

Note

In the first test case the board after the move from the hole with 7

stones will look like 1 2 2 0 0 0 0 0 0 0 1 1 1 1. Then the player collects the even numbers and ends up with a score equal to 4

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll a[20];
ll b[20];
void init()
{
    for(int i = 0; i < 14; i++)
        b[i] = a[i];
}
int main()
{
    for(int i = 0; i < 14; i++)
        scanf("%lld", &a[i]);
    ll ans = 0, sum = 0;
    for(int i = 0; i < 14; i++)
    {
        init();
        sum = 0;
        if(a[i])
        {
            b[i] = 0;
            for(int j = 0; j < 14; j++)
                b[j] += a[i] / 14;
            for(int t = 0; t < a[i] - a[i]/14 * 14; t++)
                b[(i + t + 1) % 14] ++;
            for(int k = 0; k < 14; k++)
                if(b[k] % 2 == 0)
                    sum += b[k];
            ans = max(sum, ans);
        }
    }
    printf("%lld\n", ans);
    return 0;
}