Problem:
Score : 400 points

Problem Statement

There is a bar of chocolate with a height of H blocks and a width of W blocks.
Snuke is dividing this bar into exactly three pieces.
He can only cut the bar along borders of blocks, and the shape of each piece must be a rectangle.

Snuke is trying to divide the bar as evenly as possible.
More specifically, he is trying to minimize S_{max} - S_{min}, where S_{max} is the area (the number of blocks contained) of the largest piece, and S_{min} is the area of the smallest piece.
Find the minimum possible value of S_{max} - S_{min}.

Constraints

2 ≤ H, W ≤ 10^5

Input

Input is given from Standard Input in the following format:

H W

Output

Print the minimum possible value of S_{max} - S_{min}.

Sample Input 1

3 5

Sample Output 1

0

In the division below, S_{max} - S_{min} = 5 - 5 = 0.

Sample Input 2

4 5

Sample Output 2

2

In the division below, S_{max} - S_{min} = 8 - 6 = 2.

Sample Input 3

5 5

Sample Output 3

4

In the division below, S_{max} - S_{min} = 10 - 6 = 4.

Sample Input 4

100000 2

Sample Output 4

1

Sample Input 5

100000 100000

Sample Output 5

50000

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INF 10000000000
#define mod 1000000007
#define X 1000000
typedef long long int ll;
ll finv[X];
ll fac[X];
ll inv[X];
ll usort(const void *a, const void *b){return *(ll*)a-*(ll*)b;}
ll dsort(const void *a, const void *b){return *(ll*)b-*(ll*)a;}
ll ssort(const void *a, const void *b){return strcmp((char*)a,(char*)b);}
ll min(ll a,ll b){
  if(a>b)return b;
  return a;
}
ll max(ll a,ll b){
  if(a>b)return a;
  return b;
}
ll z(ll a){return max(a,-a);}
ll gcd(ll a,ll b){if(!b){return a;}else{return gcd(b,a%b);}}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
ll kt(ll a){
  ll sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
ll ks(ll a){
  ll sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
void ppp(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for (int i=2;i<X;i++){
    fac[i]=fac[i-1]*i%mod;
    inv[i]=mod-inv[mod%i]*(mod/i)%mod;
    finv[i]=finv[i-1]*inv[i]%mod;
  }
}
ll nCr(int n,int k){
  return fac[n]*(finv[k]*finv[n-k]%mod)%mod;
}
int main(void){
  ll n,m;
  scanf("%lld %lld",&n,&m);
  if(n%3<1||m%3<1){
    PRN(0);
    return 0;
  }
  ll a[3],ans;
  for(ll i=1;i<n;i++){
    if(i==1)ans=max(z(a[0]-a[1]),max(z(a[1]-a[2]),z(a[0]-a[2])))+1;
    a[0]=i*m;
    a[1]=(n-i)*(m/2);
    a[2]=(n-i)*(m/2+m%2);
    ll before=ans;
    ans=min(ans,max(z(a[0]-a[1]),max(z(a[1]-a[2]),z(a[0]-a[2]))));
    if(before==ans)break;
  }
  if(n-m){
    for(ll i=1;i<m;i++){
      if(i==1)ans=max(z(a[0]-a[1]),max(z(a[1]-a[2]),z(a[0]-a[2])))+1;
      a[0]=i*n;
      a[1]=(m-i)*(n/2);
      a[2]=(m-i)*(n/2+n%2);
      ll before=ans;
      ans=min(ans,max(z(a[0]-a[1]),max(z(a[1]-a[2]),z(a[0]-a[2]))));
      if(before==ans)break;
    }
  }
  PRL(ans);
  return 0;
}