Recently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integer x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer i (1 ≤ i ≤ n) and replaces the i-th element of array ai either with ai + x or with ai - x. Please note that the operation may be applied more than once to the same position.

Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e. <image>) can reach, if Maxim would apply no more than k operations to it. Please help him in that.

Input

The first line of the input contains three integers n, k and x (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 109) — the number of elements in the array, the maximum number of operations and the number invented by Maxim, respectively.

The second line contains n integers a1, a2, ..., an (<image>) — the elements of the array found by Maxim.

Output

Print n integers b1, b2, ..., bn in the only line — the array elements after applying no more than k operations to the array. In particular, <image> should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum possible.

If there are multiple answers, print any of them.

Examples

Input

5 3 1
5 4 3 5 2


Output

5 4 3 5 -1 


Input

5 3 1
5 4 3 5 5


Output

5 4 0 5 5 


Input

5 3 1
5 4 4 5 5


Output

5 1 4 5 5 


Input

3 2 7
5 4 2


Output

5 11 -5 
#include <bits/stdc++.h>
using namespace std;
vector<long long int> fact(200005);
vector<long long int> prime_check(1e7 + 2);
vector<long long int> primes;
long long powmod(long long x, long long y, long long m) {
  long long res = 1LL;
  while (y) {
    if (y & 1) res = (res * x) % m;
    x = (x * x) % m;
    y /= 2;
  }
  return res;
}
void init_fact() {
  fact[0] = 1;
  for (long long int i = 1; i < 200005; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000009;
  }
}
long long int C(long long int n, long long int r) {
  if (n < r) return 0;
  long long int temp =
      fact[n] * powmod(fact[r], 1000000009 - 2, 1000000009) % 1000000009;
  temp *= powmod(fact[n - r], 1000000009 - 2, 1000000009);
  temp %= 1000000009;
  return temp;
}
void Sieve() {
  for (long long int i = 2; i <= 1e7 + 1; i++) {
    if (prime_check[i]) continue;
    primes.push_back(i);
    for (long long int j = 2 * i; j <= 1e7 + 1; j += i) prime_check[j] = i;
  }
}
long long int phi(long long int n) {
  long long int i, res = n;
  for (i = 2; i * i <= n; i++)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      res -= res / i;
    }
  if (n > 1) res -= res / n;
  return res;
}
long long int n, k, x;
vector<long long int> a(200005);
set<pair<long long int, long long int>> elem;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k >> x;
  long long int sign = 1;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
    elem.insert(make_pair(abs(a[i]), i));
    if (a[i] < 0) sign *= -1;
  }
  while (k > 0) {
    auto temp = *elem.begin();
    long long int ind = temp.second;
    if (a[ind] < 0) sign *= -1;
    if (sign == 1)
      a[ind] -= x;
    else
      a[ind] += x;
    if (a[ind] < 0) sign *= -1;
    elem.erase(elem.begin());
    elem.insert(make_pair(abs(a[ind]), ind));
    k--;
  }
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
  cout << "\n";
}
