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 inf = 1 << 30;
const long long INF = (long long)1 << 62;
const int mod = 1e9 + 7;
const long long maxn = 2 * 1e5 + 10;
long long n, k, x, a[maxn], sign;
set<pair<long long, long long> > s, ans;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k >> x;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] < 0LL) sign++;
    s.insert({a[i], i});
  }
  auto t = s.lower_bound(make_pair(0LL, -1LL));
  while (t->first == 0LL && k > 0) {
    long long t1 = t->first + x, t2 = t->second;
    if (sign % 2 == 0LL) {
      t1 -= 2 * x;
      sign++;
    }
    s.erase(t);
    s.insert(make_pair(t1, t2));
    k--;
    t = s.lower_bound(make_pair(0LL, -1LL));
  }
  while (k > 0LL) {
    t = s.lower_bound(make_pair(0LL, -1LL));
    long long t1, t2;
    if (sign % 2 == 0LL) {
      if (t != s.begin() && t != s.end()) {
        auto temp = prev(t);
        if (abs(temp->first) < t->first) {
          t1 = temp->first + x, t2 = temp->second;
          if (t1 >= 0LL) sign++;
          s.erase(temp);
        } else {
          t1 = t->first - x, t2 = t->second;
          if (t1 < 0LL) sign++;
          s.erase(t);
        }
      } else if (t == s.begin()) {
        t1 = t->first - x, t2 = t->second;
        if (t1 < 0LL) sign++;
        s.erase(t);
      } else {
        auto temp = prev(t);
        t1 = temp->first + x, t2 = temp->second;
        if (t1 >= 0LL) sign++;
        s.erase(temp);
      }
    } else {
      if (t != s.begin() && t != s.end()) {
        auto temp = prev(t);
        if (abs(temp->first) < t->first) {
          t1 = temp->first - x, t2 = temp->second;
          s.erase(temp);
        } else {
          t1 = t->first + x, t2 = t->second;
          s.erase(t);
        }
      } else if (t == s.begin()) {
        t1 = t->first + x, t2 = t->second;
        s.erase(t);
      } else {
        auto temp = prev(t);
        t1 = temp->first - x, t2 = temp->second;
        s.erase(temp);
      }
    }
    s.insert(make_pair(t1, t2));
    k--;
  }
  for (auto it = s.begin(); it != s.end(); ++it)
    ans.insert(make_pair(it->second, it->first));
  for (auto it = ans.begin(); it != ans.end(); ++it) cout << it->second << ' ';
  return 0;
}
