C. Block Towers(二分)

C. Block Towers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Students in a class are making towers of blocks. Each student makes a (non-zero) tower by stacking pieces lengthwise on top of each other. n of the students use pieces made of two blocks and m of the students use pieces made of three blocks.

The students don’t want to use too many blocks, but they also want to be unique, so no two students’ towers may contain the same number of blocks. Find the minimum height necessary for the tallest of the students' towers.

Input

The first line of the input contains two space-separated integers n and m (0 ≤ n, m ≤ 1 000 000, n + m > 0) — the number of students using two-block pieces and the number of students using three-block pieces, respectively.

Output

Print a single integer, denoting the minimum possible height of the tallest tower.

Examples

input

Copy

1 3

output

Copy

9

input

Copy

3 2

output

Copy

8

input

Copy

5 0

output

Copy

10

Note

In the first case, the student using two-block pieces can make a tower of height 4, and the students using three-block pieces can make towers of height 3, 6, and 9 blocks. The tallest tower has a height of 9 blocks.

In the second case, the students can make towers of heights 2, 4, and 8 with two-block pieces and towers of heights 3 and 6 with three-block pieces, for a maximum height of 8 blocks.

题意：给你n,m；让你给出n个2的倍数构成一个集合,m个3的倍数构成一个集合，求两个集合中最大的元素（两个集合中不能有相同的元素）

思路：关键在于6的倍数是在3那里还是在2那里

num1:2的倍数

num2:3的倍数

num3:6的倍数 然后可以推出上图的三个限定条件

二分一下最大的数，然后满足上面的三个条件即可

这个题能用二分的思路在于 在一段区间里找一个数，然后给出了这个数合法的判定条件

ac代码

#include<bits/stdc++.h>

using namespace std;

int n,m;

bool check(int x)
{
    int num1 = x/2;
    int num2 = x/3;
    int num3 = x/6;
    if(num1 < n)
        return false;
    if(num2 < m)
        return false;
    if(min(num3,num1-n)<m-(num2-num3) )
        return false;

    return true;

}

int main()
{
    cin>>n>>m;
    int l = 0 ,r = 1e7 ,ans = 0;
    while(l <= r)
    {
        int mid = (l+r)/2;
        if(check(mid))
        {
            ans = mid;
            r = mid-1;
        }
        else
            l = mid+1;
    }
    cout<<ans<<endl;
}