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: Accepted)

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

int main() {
  long long int i, j;
  long long int H, W;

  scanf("%lld %lld", &H, &W);

  long long int minH, minW;
  long long int tempH, tempW;
  if (H % 3 == 0) {
    printf("0");
    return 0;
  }
  if (W % 3 == 0) {
    printf("0");
    return 0;
  }

  tempH = H - H / 3;
  minH = W;
  if (W % 2 == 0) {
    if (minH > (tempH * W / 2 - W * (H - tempH))) {
      minH = (tempH * W / 2 - W * (H - tempH));
    }
  } else {
    if (minH > (tempH * (W + 1) / 2 - W * (H - tempH))) {
      minH = (tempH * (W + 1) / 2 - W * (H - tempH));
    }
  }
  tempH--;
  if (W % 2 == 0) {
    if (minH > -(tempH * W / 2 - W * (H - tempH))) {
      minH = -(tempH * W / 2 - W * (H - tempH));
    }
  } else {
    if (minH > -(tempH * (W - 1) / 2 - W * (H - tempH))) {
      minH = -(tempH * (W - 1) / 2 - W * (H - tempH));
    }
  }

  tempW = W - W / 3;
  minW = H;
  if (H % 2 == 0) {
    if (minW > (tempW * H / 2 - H * (W - tempW))) {
      minW = (tempW * H / 2 - H * (W - tempW));
    }
  } else {
    if (minW > (tempW * (H + 1) / 2 - H * (W - tempW))) {
      minW = (tempW * (H + 1) / 2 - H * (W - tempW));
    }
  }
  tempW--;
  if (H % 2 == 0) {
    if (minW > -(tempW * H / 2 - H * (W - tempW))) {
      minW = -(tempW * H / 2 - H * (W - tempW));
    }
  } else {
    if (minW > -(tempW * (H - 1) / 2 - H * (W - tempW))) {
      minW = -(tempW * (H - 1) / 2 - H * (W - tempW));
    }
  }

  if (minH < minW) {
    printf("%lld", minH);
  }
  else {
    printf("%lld", minW);
  }

  return 0;
}