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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val (H, W) = readLine()!!.split(" ").map { it.toLong() }

    // Tips: 全てを解析的に求める方針は筋が悪い 数値 * 解析 の組み合わせで実装する方針だと簡単に解ける良い例


    var ans = Long.MAX_VALUE

    fun calc(h: Long, w: Long) {
        for ( i in 1..h-1) {
            var one = i * w

            // 全て横分割ケース
            var two = ((h-i)/2) * w
            var three = (h - (h-i)/2) * w

            var listH = listOf(one, two, three)
            ans = Math.min(listH.max()!!-listH.min()!!, ans)

            // 横１縦２分割ケース
            var twoV = (h-i) * (w/2)
            var threeV = (h-i) * (w - w/2)
            var listV = listOf(one, twoV, threeV)
            ans = Math.min(listV.max()!!-listV.min()!!, ans)
        }
    }


    calc(H, W)
    // 縦横入れ替えた場合も計算
    calc(W, H)

    println(ans)

    /**
    // ２パターンの方法で行ける気がする
    if ( h%3==0L || w%3==0L) {
        println(0)
        return
    }

    // どちらか一方でも偶数の場合
    if ( h%2==0L || w%2==0L) {
        // 奇数の側でまず分割し，偶数側で残った大きい方を分割
        if( w%2 == 0L) {
            val one = h/2 * w
            val two = (h/2 + 1) * (w/2)

            println(Math.abs(one-two))
            return
        } else {
            val one = w/2 * h
            val two = (w/2 + 1) * (h/2)

            println(Math.abs(one-two))
            return
        }

    }

    // 両方とも奇数の場合
    val one = h/2 * w
    val two = (h/2 + 1) * (w/2)

    println(Math.abs(one-two))
    **/
}