CodeForces 651 A Joysticks

A. Joysticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

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

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
long long int dp[205][205];
int n,m;
long long int dfs(int x,int y)
{
    //cout<<x<<" "<<y<<endl;
    if(dp[x][y]!=-1)
        return dp[x][y];
    if(x-2<0&&y-2>=0)
       dp[x][y]=dfs(x+1,y-2)+1;
    else if(x-2>=0&&y-2<0)
       dp[x][y]=dfs(x-2,y+1)+1;
    else
    {
       dp[x][y]=max(dfs(x+1,y-2),dfs(x-2,y+1));
       dp[x][y]++;
    }

    return dp[x][y];
}
int main()
{
    memset(dp,-1,sizeof(dp));
    for(int i=1;i<=100;i++)
        dp[i][0]=0,dp[0][i]=0;
    dp[1][1]=0;
    scanf("%d%d",&n,&m);
    printf("%lld\n",dfs(n,m));
    return 0;
}