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;
long long i, j, k, l, m, n, s, a[211111], b[222222], c[211], d, e, t, x;
string p, q[111], r, w[111];
pair<long long, long long> y[211111];
vector<long long> u[1111], uu;
multiset<pair<long long, long long> > SET;
multiset<pair<long long, long long> >::iterator it;
int main() {
  cin >> n >> k >> m;
  for (i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
    c[i] = 1;
    if (a[i] < 0) {
      c[i] = -1;
      a[i] = -a[i];
      t++;
    }
    y[i].first = a[i];
    y[i].second = i;
  }
  y[n].first = 111111111111111111;
  y[n].second = n;
  sort(y, y + n);
  if (y[0].first > k * m && t % 2 == 0) {
    y[0].first -= k * m;
    for (i = 0; i < n; i++) SET.insert(make_pair(y[i].first, y[i].second));
  } else {
    if (t % 2 == 0) {
      d = y[0].first / m;
      k -= d;
      y[0].first %= m;
      if (k > 0) {
        y[0].first -= m;
        c[y[0].second] *= -1;
        y[0].first *= -1;
        k--;
      }
    }
    for (i = 0; i < n; i++) SET.insert(make_pair(y[i].first, y[i].second));
    while (k--) {
      pair<long long, long long> pp;
      it = SET.begin();
      pp = *it;
      SET.erase(pp);
      pp.first += m;
      SET.insert(pp);
    }
  }
  for (it = SET.begin(); it != SET.end(); it++) a[it->second] = it->first;
  for (i = 0; i < n; i++) printf("%I64d ", a[i] * c[i]);
}
