You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.

You can apply the following operation to this permutation, any number of times (possibly zero):

* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.



Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.

Input

The input is given from Standard Input in the following format:


N K
P_1 P_2 ... P_N


Output

Print the lexicographically smallest permutation that can be obtained.

Examples

Input

4 2
4 2 3 1


Output

2
1
4
3


Input

5 1
5 4 3 2 1


Output

1
2
3
4
5


Input

8 3
4 5 7 8 3 1 2 6


Output

1
2
6
7
5
3
4
8
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 50;
int n, k, a[N], pos[N], deg[N], f[N];
vector<int> g[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int i = 1; i <= n; ++i) pos[a[i]] = i;
  set<int> L, R;
  for (int i = 1; i <= k; ++i) R.insert(a[i]);
  for (int i = 1; i <= n; ++i) {
    auto p = L.upper_bound(a[i]);
    if (p != L.end()) {
      g[pos[*p]].push_back(i);
      ++deg[i];
    }
    p = R.upper_bound(a[i]);
    if (p != R.end()) {
      g[pos[*p]].push_back(i);
      ++deg[i];
    }
    L.insert(a[i]), R.erase(a[i]);
    if (i - k + 1 > 0) L.erase(a[i - k + 1]);
    if (i + k <= n) R.insert(a[i + k]);
  }
  priority_queue<int> q;
  for (int i = 1; i <= n; ++i)
    if (!deg[i]) q.push(i);
  for (int i = n; i >= 1; --i) {
    int u = q.top();
    q.pop();
    f[i] = u;
    for (int v : g[u]) {
      if (!--deg[v]) q.push(v);
    }
  }
  for (int i = 1; i <= n; ++i) a[f[i]] = i;
  for (int i = 1; i <= n; ++i) printf("%d\n", a[i]);
}
