Codeforces Round #469 (Div. 2) A. Left-handers, Right-handers and Ambidexters（水）

题目链接：点击打开链接

A. Left-handers, Right-handers and Ambidexters

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You are at a water bowling training. There are l people who play with their left hand, r people,
who play with their right hand, and aambidexters, who can play with left or right hand.

The coach decided to form a team of even number of players, exactly half of the players should play with their right hand, and exactly half of the players should play with their left hand. One player should use only on of his hands.

Ambidexters play as well with their right hand as with their left hand. In the team, an ambidexter can play with their left hand, or with their right hand.

Please find the maximum possible size of the team, where equal number of players use their left and right hands, respectively.

Input

The only line contains three integers l, r and a (0 ≤ l, r, a ≤ 100)
— the number of left-handers, the number of right-handers and the number of ambidexters at the training.

Output

Print a single even integer — the maximum number of players in the team. It is possible that the team can only have zer
4000o number of players.

Examples

input

Copy

1 4 2

output

6

input

Copy

5 5 5

output

14

input

Copy

0 2 0

output

0

Note

In the first example you can form a team of 6 players. You should take the only left-hander and two ambidexters to play with left hand, and
three right-handers to play with right hand. The only person left can't be taken into the team.

In the second example you can form a team of 14 people. You have to take all five left-handers, all five right-handers, two ambidexters to play
with left hand and two ambidexters to play with right hand.

题意：你有三种人，分别为擅长使用左手的人，擅长使用右手的人，和不仅擅长使用左手还擅长是由右手的人（天选之人？）。然后你要从这个三种人组建一直队伍。队伍的要求为：使用左手和使用右手的人一样多。求队伍的最大人数。

题解：把天选之人。分配给使用单手人数少的。然后单手数量少的乘2就行了。

具体实现看代码：

#include<iostream>
#include<algorithm>
using namespace std;

int main(){
int l,r,a;
cin >> l >> r >> a;
while(a){
int maxx = max(l,r);
int minn = min(l,r);
minn++;
a--;
l = maxx;
r = minn;
}
cout << min(l,r) * 2 <<endl;
}