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;
inline int _sign(int x) { return (x < 0 ? -1 : 1); }
struct Numb {
  long long val;
  int sign, id;
  Numb() {}
  Numb(long long v, int s, int i) {
    val = v;
    sign = s;
    id = i;
  }
};
inline bool operator<(const Numb& a, const Numb& b) { return a.val < b.val; }
void solve() {
  int n, k;
  long long x;
  cin >> n >> k >> x;
  vector<long long> arr(n);
  int cntz = 0, neg = 0;
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &arr[i]);
    cntz += (arr[i] == 0);
    neg += (arr[i] < 0);
  }
  if (cntz > k) {
    for (auto z : arr) {
      printf("%I64d ", z);
    }
    return;
  }
  int limit = k;
  for (int i = 0; i < n; i++) {
    if (arr[i] == 0 && (cntz > 1 || neg % 2)) {
      cntz--;
      limit--;
      arr[i] = x;
    } else if (arr[i] == 0 && cntz == 1 && neg % 2 == 0) {
      cntz--;
      limit--;
      arr[i] = -x;
      neg++;
    }
  }
  if (limit == 0) {
    for (auto z : arr) {
      printf("%I64d ", z);
    }
    return;
  }
  multiset<Numb> nums;
  int i = 0;
  for (auto x : arr) {
    nums.insert(Numb(abs(x), _sign(x), i++));
  }
  Numb cur;
  long long real_cur = 0;
  long long kf;
  if (neg % 2 == 0) {
    cur = *(nums.begin());
    nums.erase(nums.begin());
    kf = (cur.val / x) + 1ll;
    if (limit >= kf) {
      real_cur = abs(abs(cur.val) - (x * kf));
      arr[cur.id] = abs(real_cur) * (-cur.sign);
      nums.insert(Numb(abs(real_cur), -cur.sign, cur.id));
      limit -= kf;
    } else {
      real_cur = abs(abs(cur.val) - (x * limit));
      arr[cur.id] = abs(real_cur) * (cur.sign);
      nums.insert(Numb(abs(real_cur), cur.sign, cur.id));
      limit = 0;
    }
  }
  for (int i = 0; i < limit; i++) {
    cur = *(nums.begin());
    nums.erase(nums.begin());
    real_cur = cur.val + x;
    arr[cur.id] = real_cur * cur.sign;
    nums.insert(Numb(real_cur, cur.sign, cur.id));
  }
  for (auto z : arr) {
    printf("%I64d ", z);
  }
}
int main() {
  if (0) {
    freopen("mushrooms.in", "r", stdin);
    freopen("mushrooms.out", "w", stdout);
  }
  solve();
  cerr << endl << "Time: " << clock() / 1000.0 << " ms";
  return 0;
}
