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;
long long int count(long long int n) {
  long long int c = 0;
  while (n != 0) {
    c++;
    n = n / 10;
  }
  return c;
}
int isPrime(long long int n) {
  if (n <= 1) return 0;
  for (long long int i = 2; i < n; i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int modFact(long long int n, long long int p) {
  if (n >= p) return 0;
  long long int result = 1;
  for (long long int i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
bool cmp(pair<string, pair<long long int, long long int>>& a,
         pair<string, pair<long long int, long long int>>& b) {
  return a.second < b.second;
}
long long int getMaxOccuringChar(string str) {
  long long int count[256] = {0};
  long long int len = str.length();
  long long int max = 0;
  char result;
  for (long long int i = 0; i < len; i++) {
    count[str[i]]++;
    if (max < count[str[i]]) {
      max = count[str[i]];
      result = str[i];
    }
  }
  return count[result];
}
long long int printPairs(long long int arr[], long long int n,
                         long long int sum) {
  unordered_map<long long int, long long int> m;
  long long int x = 0;
  for (long long int i = 0; i < n; i++) {
    long long int rem = sum - arr[i];
    if (m.find(rem) != m.end()) {
      long long int count = m[rem];
      x = x + count;
    }
    m[arr[i]]++;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, x;
  cin >> n >> k >> x;
  set<pair<long long int, long long int>> v;
  long long int m = 1;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
    v.insert(make_pair(abs(a[i]), i));
    if (a[i] < 0) m = m * -1;
  }
  while (k--) {
    pair<long long int, long long int> p = *v.begin();
    v.erase(p);
    long long int c = p.second;
    if (a[c] < 0) m = m * -1;
    if (m == 1) {
      a[c] = a[c] - x;
    } else
      a[c] = a[c] + x;
    if (a[c] < 0) m = m * -1;
    v.insert(make_pair(abs(a[c]), c));
  }
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}
