Codeforces 626C

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

1 3

output

input

3 2

output

input

5 0

output

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.

题目大意：n个人两层两层的叠塔，m个人三层三层的叠塔，每个人塔的高度要唯一，问最小的情况所有塔中最大的塔高度

ac代码

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n,m;
 4 bool check(int x)
 5 {
 6     int num1=x/2;
 7     int num2=x/3;
 8     int num3=x/6;
 9     if(num1<n)
10         return false;
11     if(num2<m)
12         return false;
13     if(n + m > num1 + num2 - num3)
14         return false;
15     return true;
16 }
17 int main()
18 {
19     scanf("%d%d",&n,&m);
20     int l=0,r=1e7,ans=0;
21     while(l<=r)
22     {
23         int mid=(l+r)/2;
24         if(check(mid))
25         {
26             ans=mid;
27             r=mid-1;
28         }
29         else l=mid+1;
30     }
31     cout<<ans<<endl;
32     return 0;
33 }