codeforces 651A Joysticks

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A. Joysticks

time limit per test

memory limit per test

input

output

a1 percent and second one is charged at a2

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.

more than 100.

Input

a1 and a2 (1 ≤ a1, a2), the initial charge level of first and second joystick respectively.

Output

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

Examples

input

3 5

output

6

input

4 4

output

5

Note

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.

模拟这个过程，让电量少的+1，另一个-2，电量不足时就结束了！注意1和1，输出0

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int a,b,m=0;
    scanf("%d%d",&a,&b);
    if(a==1&&b==1)
    {
        printf("0\n");
        return 0;
    }
    while(1)
    {
        if(a<=0||b<=0)
            break;
        if(a<=b)
        {
            a+=1;
            b-=2;
        }
        else//a>b
        {
            a-=2;
            b+=1;
        }
        m++;
    }
    printf("%d\n",m);
    return 0;
}