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;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
template <class T>
inline void smin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void smax(T &a, T b) {
  if (a < b) a = b;
}
priority_queue<pair<long long, int>, vector<pair<long long, int> >,
               greater<pair<long long, int> > >
    pq;
long long a[222222];
int sgn[222222];
int main() {
  int N, K, cnt, i, id, u;
  long long first, mn, r;
  scanf("%d%d%I64d", &N, &K, &first);
  for (i = 0; i < N; i++) scanf("%I64d", a + i);
  cnt = 0;
  for (i = 0; i < N; i++) {
    if (a[i] < 0) cnt++;
  }
  if (cnt % 2 == 0) {
    mn = 0x3f3f3f3f3f3f3f3fLL;
    for (i = 0; i < N; i++) {
      if (abs(a[i]) < mn) {
        mn = abs(a[i]);
        id = i;
      }
    }
    if (abs(a[id]) < first * K) {
      r = (abs(a[id]) + first) / first;
      if (a[id] < 0)
        a[id] += r * first;
      else
        a[id] -= r * first;
      K -= r;
      for (i = 0; i < N; i++) {
        if (a[i] >= 0)
          sgn[i] = 1;
        else
          sgn[i] = -1;
        a[i] = abs(a[i]);
      }
      for (i = 0; i < N; i++) pq.push(pair<long long, int>(a[i], i));
      for (i = 0; i < K; i++) {
        u = pq.top().second;
        pq.pop();
        a[u] += first;
        pq.push(pair<long long, int>(a[u], u));
      }
      for (i = 0; i < N; i++) a[i] *= sgn[i];
    } else {
      if (a[id] < 0)
        a[id] += first * K;
      else
        a[id] -= first * K;
    }
  } else {
    for (i = 0; i < N; i++) {
      if (a[i] >= 0)
        sgn[i] = 1;
      else
        sgn[i] = -1;
      a[i] = abs(a[i]);
    }
    for (i = 0; i < N; i++) pq.push(pair<long long, int>(a[i], i));
    for (i = 0; i < K; i++) {
      u = pq.top().second;
      pq.pop();
      a[u] += first;
      pq.push(pair<long long, int>(a[u], u));
    }
    for (i = 0; i < N; i++) a[i] *= sgn[i];
  }
  for (i = 0; i < N; i++) printf("%I64d ", a[i]);
  puts("");
  return 0;
}
