Problem:
Score : 300 points

Problem Statement

Takahashi, who lives on the number line, is now at coordinate X. He will make exactly K moves of distance D in the positive or negative direction.

More specifically, in one move, he can go from coordinate x to x + D or x - D.

He wants to make K moves so that the absolute value of the coordinate of the destination will be the smallest possible.

Find the minimum possible absolute value of the coordinate of the destination.

Constraints

-10^{15} \leq X \leq 10^{15}

1 \leq K \leq 10^{15}

1 \leq D \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X K D

Output

Print the minimum possible absolute value of the coordinate of the destination.

Sample Input 1

6 2 4

Sample Output 1

2

Takahashi is now at coordinate 6. It is optimal to make the following moves:

Move from coordinate 6 to (6 - 4 =) 2.

Move from coordinate 2 to (2 - 4 =) -2.

Here, the absolute value of the coordinate of the destination is 2, and we cannot make it smaller.

Sample Input 2

7 4 3

Sample Output 2

1

Takahashi is now at coordinate 7. It is optimal to make, for example, the following moves:

Move from coordinate 7 to 4.

Move from coordinate 4 to 7.

Move from coordinate 7 to 4.

Move from coordinate 4 to 1.

Here, the absolute value of the coordinate of the destination is 1, and we cannot make it smaller.

Sample Input 3

10 1 2

Sample Output 3

8

Sample Input 4

1000000000000000 1000000000000000 1000000000000000

Sample Output 4

1000000000000000

The answer can be enormous.

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

let lines = [];
let reader = require('readline').createInterface({
    input: process.stdin,
    output: process.stdout
});

// 標準入力の読み込み
reader.on('line', function(line) {
    lines.push(line);
});

reader.on('close', function() {

    // 処理の記述
    let [x, k, d] = lines[0].split(' ');
    x = parseInt(x);
    k = parseInt(k);
    d = parseInt(d);

    let ans = Math.abs(x);

    // xの絶対値がk * dより大きいとき
    if (Math.abs(x) > k * d) {
        if (x < 0) {
            ans = Math.abs(x + k * d);
        } else {
            ans = Math.abs(x - k * d);
        }
    } else {
        if (x < 0) {
            let cnt = 0;
            while( Math.abs(x) > d ) {
                x += d;
                cnt++;
            }

            ans = (k - cnt) % 2 == 0 ? Math.abs(x) : Math.abs(x + d);

        } else {
            let cnt = 0;
            while( Math.abs(x) > d ) {
                x -= d;
                cnt++;
            }
            ans = (k - cnt) % 2 == 0 ? Math.abs(x) : Math.abs(x - d);
        }
    }

    // 表示
    console.log(ans);
});