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;
const int mod = 100000007;
const int M = 300007;
const long double pi = acos(-1);
long long dp[M], i, k, m, n, cnt1, cn2, nt3, th4, ans5, l, r, Orto;
int main() {
  ios_base::sync_with_stdio(false);
  scanf("%I64d%I64d%I64d", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &dp[i]);
    if (dp[i] < 0) cnt1++;
    if (dp[i] == 0) cn2++;
  }
  if (m < cn2) {
    dp[1] = dp[1] + 1LL * m * k;
    for (i = 1; i <= n; i++) printf("%I64d ", dp[i]);
    return 0;
  }
  m = m - cn2;
  if (cnt1 & 1) {
    for (i = 1; i <= n; i++)
      if (dp[i] == 0) dp[i] = k;
  } else {
    for (i = 1; i <= n; i++)
      if (dp[i] == 0)
        if (!(cnt1 & 1))
          dp[i] = -k, cnt1++;
        else
          dp[i] = k;
  }
  if (!(cnt1 & 1)) {
    nt3 = abs(dp[1]) / k + 1, th4 = 1;
    for (i = 2; i <= n; i++)
      if ((abs(dp[i]) / k + 1 < nt3) ||
          (abs(dp[i]) / k + 1 == nt3) &&
              (abs(1LL * k * (abs(dp[i]) / k + 1) - abs(dp[i])) >
               abs(1LL * k * (abs(dp[th4]) / k + 1) - abs(dp[th4]))))
        nt3 = abs(dp[i]) / k + 1, th4 = i;
    nt3 = min(m, nt3);
    m = m - nt3;
    if (dp[th4] > 0)
      dp[th4] = dp[th4] - 1LL * k * nt3;
    else
      dp[th4] = dp[th4] + 1LL * k * nt3;
    if (!m) {
      for (i = 1; i <= n; i++) printf("%I64d ", dp[i]);
      return 0;
    }
  }
  l = 0, r = 200000000000000LL;
  while (l < r) {
    Orto = (l + r) / 2;
    ans5 = 0;
    for (i = 1; i <= n; i++)
      if (Orto > abs(dp[i])) ans5 = ans5 + (Orto - abs(dp[i])) / k;
    if (ans5 >= m)
      r = Orto;
    else
      l = Orto + 1;
  }
  for (int i = 1; i <= n; i++)
    if (l > abs(dp[i]))
      if ((l - abs(dp[i])) % k) {
        m = m - (l - abs(dp[i])) / k;
        if (dp[i] > 0)
          dp[i] = dp[i] + 1LL * k * ((l - abs(dp[i])) / k);
        else
          dp[i] = dp[i] - 1LL * k * ((l - abs(dp[i])) / k);
      } else {
        m = m - (l - abs(dp[i])) / k + 1;
        if (dp[i] > 0)
          dp[i] = dp[i] + 1LL * k * ((l - abs(dp[i])) / k - 1);
        else
          dp[i] = dp[i] - 1LL * k * ((l - abs(dp[i])) / k - 1);
      }
  for (int i = 1; i <= n; i++)
    if ((abs(dp[i]) == l - k) && (m > 0)) {
      m--;
      if (dp[i] > 0)
        dp[i] = dp[i] + k;
      else
        dp[i] = dp[i] - k;
    }
  for (i = 1; i <= n; i++) cout << dp[i] << " ";
  return 0;
}
