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 GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
bool chk(string first, string second) {
  string t1 = first + second;
  string t2 = second + first;
  return t1 < t2;
}
struct sort_pred {
  bool operator()(const pair<int, int> &left, const pair<int, int> &right) {
    return left.second < right.second;
  }
};
long long POW(long long Base, long long Exp) {
  long long y, ret = 1;
  y = Base;
  while (Exp) {
    if (Exp & 1) ret = (ret * y) % 1000000007;
    y = (y * y) % 1000000007;
    Exp /= 2;
  }
  return ret % 1000000007;
}
vector<long long> A, B, C, D, Res, Mark;
string str2, str1, s1, s2;
set<string> st;
stack<long long> chlo;
vector<pair<long long, long long> > Rec;
string ItoS(long long n) {
  long long num = n;
  long long tmp;
  string ans = "";
  while (num) {
    tmp = num % 10;
    ans = ans + (char)('0' + tmp);
    num /= 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int cont(long long n) {
  long long count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
set<long long> At, Bt;
set<long long>::iterator itr;
vector<string> kyu;
map<long long, long long> Reg;
map<long long, long long>::iterator it;
struct MinQ_chk {
  bool operator()(long long x, long long y) { return abs(A[x]) > abs(A[y]); }
};
priority_queue<long long, vector<long long>, MinQ_chk> MinQ;
int main() {
  long long i, j, n, k, m, x, y, z, t, tmp1, tmp2, tmp3, ind, Max, d1, d2;
  cin >> n >> k >> x;
  A.resize(n);
  for (i = 0; i < n; i++) cin >> A[i];
  if (n == 1) {
    tmp1 = A[0] - k * x;
    cout << tmp1 << endl;
    return 0;
  }
  y = 0;
  for (i = 0; i < n; i++) {
    MinQ.push(i);
    if (A[i] < 0) y++;
  }
  while (k--) {
    i = MinQ.top();
    MinQ.pop();
    if (y & 1) {
      if (A[i] >= 0)
        A[i] += x;
      else
        A[i] -= x;
    } else {
      if (A[i] < 0) {
        y--;
        A[i] += x;
      } else
        A[i] -= x;
      if (A[i] < 0) y++;
    }
    MinQ.push(i);
  }
  for (i = 0; i < n; i++) {
    cout << A[i] << " ";
  }
  return 0;
}
