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: Accepted)

#include <iostream>
#ifdef MBP
#include "stdc++.h"
#else
#include <bits/stdc++.h>
#endif

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define REP(i, n) for (long long int i = 0, i##_len = (n); i < i##_len; ++i)
#define REP1(i, n) for (long long int i = 1, i##_len = (n); i <= i##_len; ++i)
#define REP_R(i, n) for (long long int i = (n - 1); i >= 0; --i)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((long long int)(x).size())
#define BIT(n) (1LL << (n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

//#define DEBUG

#ifdef DEBUG
#define OUT(x) std::cout << (x) << std::endl
#define OUT2(x, y) std::cout << (x) << "," << (y) << std::endl
#define OUT3(x, y, z) std::cout << (x) << "," << (y) << "," << (z) << std::endl
#define OUTSP(x) std::cout << (x) << " ";
#define OUTLF std::cout << std::endl
#define OUTPA(p) OUT2(p.first, p.second)
#else
#define OUT(x) NULL
#define OUT2(x, y) NULL
#define OUT3(x, y, z) NULL
#define OUTPA(p) NULL
#define OUTSP(p) NULL
#define OUTLF NULL
#endif

#define INF INT_MAX
#define INFLL LLONG_MAX
#define PB push_back
#define MOD 1000000007LL

#define LSB(x) __builtin_ctzll(x)
#define MSB(x) 64LL - __builtin_clzll(x) - 1LL
#define MSBINT(x) 32LL - __builtin_clz(x)
#define BITCNT(x) __builtin_popcountll(x)

using namespace std;
using uif = uint_fast64_t;
using ll = long long int;
using vec = vector<ll>;
using mat = vector<vector<ll>>;
using pa = pair<ll, ll>;
using tTree = __gnu_pbds::tree<ll, __gnu_pbds::null_type, less<ll>,
                               __gnu_pbds::rb_tree_tag,
                               __gnu_pbds::tree_order_statistics_node_update>;

template <class T> inline void input(T &arr) {
  ll n = arr.size();
  REP(i, n)
  cin >> arr[i];
}
template <class T> inline void input2(T &arr1, T &arr2) {
  ll n = arr1.size();
  REP(i, n)
  cin >> arr1[i] >> arr2[i];
}
template <class T> inline void inputPA(T &arr) {
  ll n = arr.size();
  REP(i, n)
  cin >> arr[i].first >> arr[i].second;
}
template <class T> inline void outArr(T &arr) {
  for (auto x : arr) {
    OUTSP(x);
  }
  OUTLF;
}

template <class T> using PQ = priority_queue<T>;
template <class T> using PQG = priority_queue<T, vector<T>, greater<T>>;

const ll dx[4] = {1, 0, -1, 0};
const ll dy[4] = {0, 1, 0, -1};
const ll dx8[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy8[8] = {0, 1, 1, 1, 0, -1, -1, -1};

template <class T> bool chmax(T &, const T &);
template <class T> bool chmin(T &, const T &);

ll qp(ll a, ll b);
ll qp(ll a, ll b, ll mo);

ll mod_pow(ll, ll);
ll mod_add(ll, ll);
ll mod_mul(ll, ll);
ll mod_div(ll, ll);
void make_fact(ll);
void make_invfact(ll);
void init_fact();

ll combination(ll, ll);

#define FACSIZE 200002LL

ll invfac[FACSIZE];
ll fac[FACSIZE];

int main(void) {
  ll x, k, d;
  cin >> x >> k >> d;

  if (abs(x) / d > k) {
    cout << abs(x) - k * d << endl;
    return 0;
  } else {
    ll base = abs(x) / d;
    ll rest = k - base;
    if (rest % 2 == 0) {
      cout << abs(x) % d << endl;
    } else {
      cout << d - (abs(x) % d) << endl;
    }
  }

  return 0;
}

template <class T> bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}

ll qp(ll a, ll b) {
  ll ans = 1LL;
  do {
    if (b & 1LL)
      ans = 1LL * mod_mul(ans, a) % MOD;
    a = 1LL * mod_mul(a, a) % MOD;
  } while (b >>= 1LL);
  return ans;
}
ll qp(ll a, ll b, ll mo) {
  ll ans = 1LL;
  do {
    if (b & 1LL)
      ans = 1LL * (ans % mo) * (a % mo);
    a = 1LL * (a % mo) * (a % mo);
  } while (b >>= 1LL);
  return ans;
}

ll mod_pow(ll x, ll r) {
  if (r == 0)
    return 1;
  else if (r == 1)
    return x % MOD;
  else if (r % 2 == 0) {
    ll t = mod_pow(x, r / 2) % MOD;
    return t * t % MOD;
  } else {
    ll t = mod_pow(x, r / 2) % MOD;
    ll k = t * t % MOD;
    return k * x % MOD;
  }
}
ll mod_add(ll a, ll b) { return (a + b) % MOD; }

ll mod_mul(ll a, ll b) { return (a * b) % MOD; }

ll mod_div(ll a, ll b) { return mod_mul(a, mod_pow(b, MOD - 2)); }

void make_fact(ll n) {
  fac[0] = 1;
  REP(i, n) { fac[i + 1] = mod_mul(fac[i], i + 1); }
}

void make_invfact(ll n) {
  invfac[n] = mod_pow(fac[n], MOD - 2);
  REP_R(i, n) { invfac[i] = mod_mul(invfac[i + 1], i + 1); }
}

void init_fact() {
  make_fact(FACSIZE - 1);
  make_invfact(FACSIZE - 1);
}

ll combination(ll n, ll c) {

  return mod_mul(fac[n], mod_mul(invfac[c], invfac[n - c]));
}
///////////////fact
/*
  init_fact();
  cout << fac[5] << endl;
  cout << invfac[5] << endl;
  cout << combination(8, 4) << endl;
  cout << combination(31, 8) << endl;
  //*/
///////////////////