題目：

Fox Ciel has some flowers: r red flowers, g green flowers and b blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:

• To make a "red bouquet", it needs 3 red flowers.
• To make a "green bouquet", it needs 3 green flowers.
• To make a "blue bouquet", it needs 3 blue flowers.
• To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.

Help Fox Ciel to find the maximal number of bouquets she can make.

Input

The first line contains three integers r, g and b (0 ≤ r, g, b ≤ 109) — the number of red, green and blue flowers.

Output

Print the maximal number of bouquets Fox Ciel can make.

Examples

Input

3 6 9

Output

6

Input

4 4 4

Output

4

Input

0 0 0

Output

0

Note

In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.

In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.

解題報告：  入眼你會發現是道比較水的題目，整體解法不多解釋，但是存在坑點，因為有一個情況就是6 8 8，最多應該是7朵花，6（3+1+1+1，3*2+1+1，3*2+1+1），，所以在整體的局勢下，需要增加一種判斷。

ac程式碼：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long ll;

const int  maxn=1e5+100;

int main()
{
	int a,b,c,d,e,f;
	while(scanf("%d%d%d",&a,&b,&c)!=EOF)
	{
		int ans=0;
		ans=min(a,min(b,c));
		if(ans==0)
		{
			cout<<a/3+b/3+c/3<<endl;
			continue;
		}
		a-=ans;
		b-=ans;
		c-=ans;
		ans+=a/3+b/3+c/3;
		if(a%3+b%3+c%3==4)
			ans++;//補貪心的不足 
		printf("%d\n",ans);
	}
	return 0;
 }