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 dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dy[] = {0, -1, -1, -1, 0, 1, 1, 1};
long long int dxx[] = {1, -1, 0, 0};
long long int dyy[] = {0, 0, -1, 1};
bool isvalid(long long int x, long long int y, long long int n,
             long long int m) {
  return (x >= 0 && x < n && y >= 0 && y < m);
}
long long int modexpo(long long int x, long long int y) {
  if (y == 0) return 1;
  if (y % 2) {
    long long int viky = modexpo(x, y / 2);
    return (((x * viky) % 1000000000) * viky) % 1000000000;
  } else {
    long long int viky = modexpo(x, y / 2);
    return (viky * viky) % 1000000000;
  }
}
int main() {
  long long int t, i, j, p, q, k, n, l, r, m, x;
  scanf("%lld", &n);
  scanf("%lld", &k);
  scanf("%lld", &x);
  long long int ap[n + 11];
  long long int zero = 0, neg = 0;
  long long int chek = 0;
  for (i = 0; i < n; i++) {
    scanf("%lld", &ap[i]);
    if (ap[i] == 0) {
      zero++;
    } else if (ap[i] < 0) {
      neg++;
    }
  }
  if (zero > k) {
    printf("%lld ", ap[0] - (k * x));
    for (i = 1; i < n; i++) {
      printf("%lld ", ap[i]);
    }
    return 0;
  } else {
    if (neg % 2) {
      for (i = 0; i < n; i++) {
        if (ap[i] == 0) {
          ap[i] += x;
          k--;
        }
      }
    } else {
      for (i = 0; i < n; i++) {
        if (ap[i] == 0) {
          ap[i] -= x;
          k--;
          neg++;
          break;
        }
      }
      for (i = 0; i < n; i++) {
        if (ap[i] == 0) {
          ap[i] += x;
          k--;
          break;
        }
      }
    }
    if (neg % 2) {
      chek = 1;
    }
  }
  long long int mini = (long long int)1e18;
  if (!chek) {
    for (i = 0; i < n; i++) {
      mini = min(mini, abs(ap[i]));
    }
    for (i = 0; i < n; i++) {
      if (abs(ap[i]) == mini) {
        p = abs(ap[i]);
        if (ap[i] > 0) {
          while (k--) {
            p = p - x;
            if (p < 0) {
              break;
            }
          }
        } else {
          p = ap[i];
          while (k--) {
            p = p + x;
            if (p > 0) {
              break;
            }
          }
        }
        ap[i] = p;
        break;
      }
    }
  }
  priority_queue<pair<long long int, pair<long long int, long long int> > > pq;
  for (i = 0; i < n; i++) {
    if (ap[i] < 0)
      pq.push(make_pair(-abs(ap[i]), make_pair(i, 0)));
    else
      pq.push(make_pair(-ap[i], make_pair(i, 1)));
  }
  while (k > 0) {
    k--;
    long long int val = -pq.top().first;
    long long int sign = pq.top().second.second;
    long long int indx = pq.top().second.first;
    pq.pop();
    if (sign) {
      val = val + x;
      pq.push(make_pair(-val, make_pair(indx, 1)));
    } else {
      val = -val - x;
      pq.push(make_pair(val, make_pair(indx, 0)));
    }
  }
  long long int ans[n + 11];
  while (!pq.empty()) {
    if (pq.top().second.second)
      ans[pq.top().second.first] = -pq.top().first;
    else
      ans[pq.top().second.first] = pq.top().first;
    pq.pop();
  }
  for (i = 0; i < n; i++) {
    printf("%lld ", ans[i]);
  }
  return 0;
}
