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;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
int N, K, X;
priority_queue<pair<long long, int>, vector<pair<long long, int> >,
               greater<pair<long long, int> > >
    P, Z;
priority_queue<pair<long long, int> > M;
long long ans[200111];
int main() {
  scanf("%d%d%d", &N, &K, &X);
  for (int i = 0, i_len = (N); i < i_len; ++i) {
    int x;
    scanf("%d", &x);
    if (x > 0)
      P.push(pair<long long, int>(x, i));
    else if (x < 0)
      M.push(pair<long long, int>(x, i));
    else
      Z.push(pair<long long, int>(x, i));
  }
  while (K > 0 && M.size() % 2 == 0) {
    pair<long long, int> o;
    K--;
    if (Z.size()) {
      o = Z.top();
      Z.pop();
      o.first -= X;
    } else if (P.size() && M.size()) {
      if (P.top().first < abs(M.top().first)) {
        o = P.top();
        P.pop();
        o.first -= X;
      } else {
        o = M.top();
        M.pop();
        o.first += X;
      }
    } else if (P.size()) {
      o = P.top();
      P.pop();
      o.first -= X;
    } else {
      o = M.top();
      M.pop();
      o.first += X;
    }
    if (o.first < 0) {
      M.push(o);
    } else if (o.first == 0) {
      Z.push(o);
    } else {
      P.push(o);
    }
  }
  while (K > 0) {
    K--;
    if (Z.size()) {
      pair<long long, int> o = Z.top();
      Z.pop();
      o.first += X;
      P.push(o);
    } else if (P.size() && M.size()) {
      if (P.top().first < abs(M.top().first)) {
        pair<long long, int> o = P.top();
        P.pop();
        o.first += X;
        P.push(o);
      } else {
        pair<long long, int> o = M.top();
        M.pop();
        o.first -= X;
        M.push(o);
      }
    } else if (P.size()) {
      pair<long long, int> o = P.top();
      P.pop();
      o.first += X;
      P.push(o);
    } else {
      pair<long long, int> o = M.top();
      M.pop();
      o.first -= X;
      M.push(o);
    }
  }
  while (P.size()) {
    pair<long long, int> o = P.top();
    P.pop();
    ans[o.second] = o.first;
  }
  while (M.size()) {
    pair<long long, int> o = M.top();
    M.pop();
    ans[o.second] = o.first;
  }
  while (Z.size()) {
    pair<long long, int> o = Z.top();
    Z.pop();
    ans[o.second] = o.first;
  }
  for (int i = 0, i_len = (N); i < i_len; ++i)
    printf("%lld%c", ans[i], i == N - 1 ? '\n' : ' ');
  return 0;
}
