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最新推荐文章于 2021-11-12 16:13:20 发布

原创最新推荐文章于 2021-11-12 16:13:20 发布 · 255 阅读

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本文探讨了编程竞赛训练中，如何将新手与经验丰富的参赛者合理分配到不同类型的团队中，以实现最大化的团队数量。提供了具体的算法实现思路，并通过示例展示了如何根据不同参与者的人数来确定最佳团队组合。

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A and B are preparing themselves for programming contests.

An important part of preparing for a competition is sharing programming knowledge from the experienced members to those who are just beginning to deal with the contests. Therefore, during the next team training A decided to make teams so that newbies are solving problems together with experienced participants.

A believes that the optimal team of three people should consist of one experienced participant and two newbies. Thus, each experienced participant can share the experience with a large number of people.

However, B believes that the optimal team should have two experienced members plus one newbie. Thus, each newbie can gain more knowledge and experience.

As a result, A and B have decided that all the teams during the training session should belong to one of the two types described above. Furthermore, they agree that the total number of teams should be as much as possible.

There are n experienced members and m newbies on the training session. Can you calculate what maximum number of teams can be formed?

Input

The first line contains two integers n and m (0 ≤ n, m ≤ 5·10⁵) — the number of experienced participants and newbies that are present at the training session.

Output

Print the maximum number of teams that can be formed.

Examples

Input

2 6

Output

Input

4 5

Output

Note

Let's represent the experienced players as XP and newbies as NB.

In the first test the teams look as follows: (XP, NB, NB), (XP, NB, NB).

In the second test sample the teams look as follows: (XP, NB, NB), (XP, NB, NB), (XP, XP, NB).

这道题有2+1和1+2两种情况可以等于3+3先去掉，所以只要找去掉的这个数和n，m的最小值，就是答案

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f;	


int main(){
	int n,m;
	while(~scanf("%d %d",&n,&m)){
		int ans;
		int xx=(n+m)/3;
		ans=min(xx,min(n,m));
	
		
		printf("%d\n",ans);
	}
	
}