Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at a₁ percent and second one is charged at a₂ percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).

Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.

Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.

The first line of the input contains two positive integers a₁ and a₂ (1 ≤ a₁, a₂ ≤ 100), the initial charge level of first and second joystick respectively.

Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.

In the first sample game lasts for 6 minute by using the following algorithm:

• at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%;
• continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%;
• at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%;
• continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%;
• at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%;
• at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.

After that the first joystick is completely discharged and the game is stopped.

做的第一道CF的题目自我鼓励一下- -|||（做了蛮久，没经验），题意：有个蛋疼的人有

俩游戏机，两者同时有电才能一直玩，但是充电头就一个，每分钟可以续（手动滑

稽）1%，另外一个不能充会掉电2%，求最大可玩时间。刚开始看出来这是道模拟题，

思路正确，就是while里的东西改了很久，规律：每次都找电最多的去玩，玩到再过一分

分钟就不能玩了，换另一个，这个拿去充电。循环不停，直到任意一个掉电2%关机为

止。这题有坑点，要是刚开始就不能玩或者只能玩一分钟，直接特盘掉，难怪我TEST7

显示超时....刚开始用a与b来模拟俩机器，发现不好使，还是数组好使。

代码：

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<queue>

#include<set>

#include<map>

#include<sstream>

#include<algorithm>

#include<cmath>

#include<cstdlib>

using namespace std;

int list[2];

int main (void)

{

	int a,b;

	int cnt;

	int index;

	while (cin>>list[0]>>list[1])

	{

		if((list[0]==1&&list[1]==2)||(list[0]==2&&list[1]==1)||(list[1]==2&&list[0]==2))//玩一分钟特判

		{

			cout<<1<<endl;

			continue;

		}

		else if(list[0]==1&&list[1]==1)//不能玩特判

		{

			cout<<0<<endl;

			continue;

		}

		if(list[0]>=list[1])

			swap(list[0],list[1]);//方便使用

		cnt=0;

		index=1;

		while (list[0]>0&&list[1]>0)

		{

			while (list[index]>2)

			{

				list[index]-=2;

				list[index==1?0:1]++;

				cnt++;

			}

			index=(index==1)?0:1;//这个完了换另一个

			if((list[0]==1&&list[1]==2)||(list[0]==2&&list[1]==1)||(list[0]==2&&list[1]==2))//循环内判断

			{

				cnt++;

				break;

			}

		}

		cout<<cnt<<endl;

	}

	return 0;

}