cf传送门：Gourmet Cat

vj传送门：Gourmet Cat

题目描述

Polycarp has a cat and his cat is a real gourmet! Dependent on a day of the week he eats certain type of food:

on Mondays, Thursdays and Sundays he eats fish food;

on Tuesdays and Saturdays he eats rabbit stew;

on other days of week he eats chicken stake.

Polycarp plans to go on a trip and already packed his backpack. His backpack contains:

a daily rations of fish food;

b daily rations of rabbit stew;

c daily rations of chicken stakes.

Polycarp has to choose such day of the week to start his trip that his cat can eat without additional food purchases as long as possible. Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.

Input

The first line of the input contains three positive integers a, b and c (1≤a,b,c≤7⋅1e8) — the number of daily rations of fish food, rabbit stew and chicken stakes in Polycarps backpack correspondingly.

Output

Print the maximum number of days the cat can eat in a trip without additional food purchases, if Polycarp chooses the day of the week to start his trip optimally.

Examples

input

2 1 1

output

4

input

3 2 2

output

7

input

1 100 1

output

3

input

30 20 10

output

39

Note

In the first example the best day for start of the trip is Sunday. In this case, during Sunday and Monday the cat will eat fish food, during Tuesday — rabbit stew and during Wednesday — chicken stake. So, after four days of the trip all food will be eaten.

In the second example Polycarp can start his trip in any day of the week. In any case there are food supplies only for one week in Polycarps backpack.

In the third example Polycarp can start his trip in any day, excluding Wednesday, Saturday and Sunday. In this case, the cat will eat three different dishes in three days. Nevertheless that after three days of a trip there will be 99 portions of rabbit stew in a backpack, can cannot eat anything in fourth day of a trip.

题意分析：

这道题大概的意思就是给定一个序列`abcacba abcacba......`

然后给定a，b，c的个数，求能得到的最大序列的长度。

观察可以发现3个a2个b2个c可以当成一组看待，然后将a，b，c通过这个原则进行缩小。缩小之后在进行判断，从第一个a开始最长的序列是多长，从第二个b开始最长序列是多长，从第三个c开始最长序列是多长，从第四个a开始最长的序列是多长。。。。。。用代码实现这部分功能就是

```
int ans=0,s=0;
for(int i=1;i<8;i++)
{
int j=i,s=0;
int A=a,B=b,C=c;
while(A>=0&&B>=0&&C>=0)
{
if(j%7==1||j%7==4||j%7==0) --A;
else if(j%7==3||j%7==5) --C;
else --B;
j++;
s++;
}
ans=max(ans,s-1);
}
```

每次计算出的s都需要减1，因为当你退出while循环的时候，a，b，c中会有一个值为-1，相当于多计算了一个字符，所以需要减一。如果不懂的话，建议手动模拟一下while循环中的内容，会有助于理解的。

这道题的代码实现

```
#include<iostream>
using namespace std;
int main()
{
int a,b,c;
cin>>a>>b>>c;
int x;
x=min(a/3,min(b/2,c/2));
a-=x*3;
b-=x*2;
c-=x*2;
int ans=0,s=0;
for(int i=1;i<8;i++)
{
int j=i,s=0;
int A=a,B=b,C=c;
while(A>=0&&B>=0&&C>=0)
{
if(j%7==1||j%7==4||j%7==0) --A;
else if(j%7==3||j%7==5) --C;
else --B;
j++;
s++;
}
ans=max(ans,s-1);
}
cout<<ans+7*x<<endl;
return 0;
}
```