Codeforces 626C. Block Towers

C. Block Towers

Students in a class is making towers of blocks. Each student makes a (Non-zero) tower is stacking pieces lengthwise on top of each of the other. N of the students use pieces made of both blocks and m of the students use pieces made of three B Locks.

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

Input

The first line of the input contains the space-separated integers n and m (0≤ n, C6>m ≤1, 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.

Examplesinput

1 3

Output

9

Input

3 2

Output

8

Input

5 0

Output

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 He Ights 3 and 6 with three-block pieces, for a maximum height of 8 blocks.

Idea: given to you is P 2 and Q 3, then this requirement of the minimum number of T->=max (2*P,3*Q); then starting with Max (2*Q,3*Q), the violent cycle determines whether the current number meets the requirements.

When the number is K, K/2 is a multiple of 2, K/3 is a multiple of 3, K/6 is a multiple of 6, wherein 6 is a multiple is to be met, because the multiples of 6 is obtained by 2, or 3, then (K/2-K/6) is a multiple of 2 and can not be removed 3;

Similarly (K/3-K/6) is a multiple of 3 but not divisible 2;x= (P (K/2-K/6)) is able to supply 6 of the number, the result <0 x=0; This means that 2 of the number is only 2 divisible by the number of the numbers are filled, so 6 is 0, when

However, there can be 2 of the number even those who can only be divisible by 2 of the number of the case, like the case of p=0 (this is because 2,3 is independent), the same x1= (P (K/3-K/6)), if X1+X==K/6, then is K;

1#include <Set>2#include <stack>3#include <queue>4#include <stdio.h>5#include <iostream>6#include <stdlib.h>7#include <string.h>8#include <math.h>9#include <algorithm>Ten#include <map> One using namespacestd; A intMainvoid) - { -     inti,j,k,p,q; the      while(SCANF ("%d%d", &p,&q)! =EOF) -     { -         intCc=2*p; -         intDd=3*Q; +         intyy=Max (CC,DD); -         intflag=0; +          for(I=YY;; i++) A         { at             if(i%2==0|| i%3==0) -             { -                 intuu=i/2; -                 intkk=i/3; -                 intzz=i/6; -                 intX1= (uu-ZZ); in                 intX2= (kk-ZZ); -x1=p-X1; tox2=q-x2; +                 if(x1<0) -                 { thex1=0; *                 } $                 if(x2<0)Panax Notoginseng                 { -X2=0; the                 }; +                 if(x1+x2==ZZ) A                 { the                      Break; +                 } -             } $         } $printf"%d\n", i); -  -     } the}

1#include <iostream>2#include <cstdio>3#include <cstring>4 intErintNintMintAintb);5 using namespacestd;6 intMain ()7 {8     intb;9scanf"%d%d",&a,&b);Ten     intR=10000000; One     intK=er (0, r,a,b); Aprintf"%d\n", k); -     return 0; - } the  - intErintNintMintAintb) - { -     intMid= (n+m)/2; +     if(n>m) -     { +         returnN; A     } at     intCnt=max (A-mid/2+mid/6,0) +max (B-mid/3+mid/6,0); -     if(Mid/6>=CNT) -     { -         returnER (n,mid-1, A, b); -     } -     Else in         returnER (mid+1, m,a,b); -}

1  while(l<=R)2     {3         intMid= (l+r) >>1;4         intRet=max (A-mid/2+mid/6,0) +max (B-mid/3+mid/6,0);5         if(Mid/6==ret)6         {7            Break;8}Else if(Mid/6>ret)9         {Tenr=mid-1; One         } A         ElseL=mid+1; -}

Codeforces 626C. Block Towers