CodeForces478 C.Table Decorations (贪心)

C.Table Decorations

You have r red, g green and b blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn’t have the same color. What maximum number t of tables can be decorated if we know number of balloons of each color?

Your task is to write a program that for given values r, g and b will find the maximum number t of tables, that can be decorated in the required manner.

Input

The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.

Output

Print a single integer t — the maximum number of tables that can be decorated in the required manner.

Examples

Input

5 4 3

Output

4

Input

1 1 1

Output

1

Input

2 3 3

Output

2

Note

In the first sample you can decorate the tables with the following balloon sets: “rgg”, “gbb”, “brr”, “rrg”, where “r”, “g” and “b” represent the red, green and blue balls, respectively.

思路：

先将输入的三种颜色从大到小排序，假设排序之后a>b>c

如果a>(b+c)*2，则答案为b+c 原理：两个a和一个b搭配或者两个a和一个c搭配(aab,aac)

如果a<=(b+c)*2，则答案为(a+b+c)/3，向下取整 原理：对于每张桌子，颜色最多的贡献两个，颜色次多的贡献一个 即答案肯定可以达到(a+b+c)/3

code：

#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> using namespace std; #define int long long signed main(){ int r,g,b; cin>>r>>g>>b; int ans=0; if(r<g)swap(r,g); if(r<b)swap(r,b); if(g<b)swap(g,b); //r>g>b if((g+b)*2<=r){ cout<<g+b<<endl; }else{ cout<<(r+g+b)/3<<endl; } return 0; }