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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef priority_queue<int> maxHeap;
typedef priority_queue<int, vector<int>, greater<int>> minHeap;
#define endl "\n"
#define int long long
#define MOD 1000000007
#define ump unordered_map
#define pb push_back
#define mp make_pair
#define pie acos(-1)
#define max(x, y) ((x > y) ? x : y)
#define min(x, y) ((x < y) ? x : y)
#define mid(s, e) ((s + (e - s) / 2))

bool test_cases;

struct precompute {
    void compute() {
    }
};

void init(bool k) {
    test_cases = k;
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}

struct test_cases {
    void test_case(int test) {
        int x, k, d;
        cin >> x >> k >> d;

        if (x == 0 and k % 2 == 0) {
            cout << 0 << endl;
            return;
        }

        if (x == 0) {
            cout << d << endl;
            return;
        }

        if (x > 0) {
            int z = x / d;
            if (z > k) {
                cout << x - k * d << endl;
                return;
            }
            int rem = k - z;
            if (rem % 2 == 0) {
                cout << x - z * d << endl;
                return;
            }
            cout << min(abs(x - z * d + d), abs(x - z * d - d)) << endl;
            return;
        } else {
            int z = abs(x / d);
            if (z > k) {
                cout << x + k * d << endl;
                return;
            }
            int rem = k - z;
            if (rem % 2 == 0) {
                cout << x + z * d << endl;
                return;
            }
            cout << min(abs(x + z * d + d), abs(x + z * d - d)) << endl;
            return;
        }
    }
};

int32_t main() {
    init(false);
    int tc;
    if (test_cases) {
        cin >> tc;
    } else {
        tc = 1;
    }
    struct precompute l;
    l.compute();
    for (int i = 0; i < tc; i++) {
        struct test_cases o;
        o.test_case(i + 1);
    }
    return 0;
};