CodeForces

A. Compote

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Nikolay has a lemons, b apples and c pears. He decided to cook a compote. According to the recipe the fruits should be in the ratio1: 2: 4. It means that for each lemon in the compote should be exactly 2 apples and exactly 4 pears. You can't crumble up, break up or cut these fruits into pieces. These fruits — lemons, apples and pears — should be put in the compote as whole fruits.

Your task is to determine the maximum total number of lemons, apples and pears from which Nikolay can cook the compote. It is possible that Nikolay can't use any fruits, in this case print 0.

Input

The first line contains the positive integer a (1 ≤ a ≤ 1000) — the number of lemons Nikolay has.

The second line contains the positive integer b (1 ≤ b ≤ 1000) — the number of apples Nikolay has.

The third line contains the positive integer c (1 ≤ c ≤ 1000) — the number of pears Nikolay has.

Output

Print the maximum total number of lemons, apples and pears from which Nikolay can cook the compote.

Examples

input

257

output

7

input

4713

output

21

input

232

output

0

Note

In the first example Nikolay can use 1 lemon, 2 apples and 4 pears, so the answer is 1 + 2 + 4 = 7.

In the second example Nikolay can use 3 lemons, 6 apples and 12 pears, so the answer is 3 + 6 + 12 = 21.

In the third example Nikolay don't have enough pears to cook any compote, so the answer is 0.

一道水题，做一份拼盘，每份需要1个柠檬，2个苹果，4个桃子。给出每种水果的数量，问最多能消耗多少水果。

从需要最少的柠檬算起，依次判断其他水果的数量是否够，若不够，执行柠檬减一，直到检测到数据可以，这时就是最大的柠檬数以及水果数。

#include<stdio.h>
#include<string.h>
#include<math.h>
int main()
{
int a,b,c,i;
while(scanf("%d%d%d",&a,&b,&c)!=EOF)
{
for(i=a;i>=0;i--)  
        if(2*i<=b&&4*i<=c)
{  
            printf("%d\n",i+2*i+4*i);  
            break;  
    }
}

        return 0;
}