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 INF = 2147483647;
const int Maxn = 500010;
int N, K, A[Maxn], P[Maxn];
int Degree[Maxn];
struct edge {
  int To, Next;
};
edge Edge[Maxn << 1];
int Start[Maxn], Used;
inline void AddEdge(int x, int y) {
  ++Degree[y];
  Edge[++Used] = (edge){y, Start[x]};
  Start[x] = Used;
  return;
}
namespace segmentTree {
int Tree[Maxn << 2];
void Build(int Index, int L, int R) {
  Tree[Index] = INF;
  if (L == R) return;
  int Mid = L + R >> 1;
  Build(Index << 1, L, Mid);
  Build(Index << 1 | 1, Mid + 1, R);
  return;
}
void Add(int Index, int L, int R, int k) {
  Tree[Index] = min(Tree[Index], k);
  if (L == R) return;
  int Mid = L + R >> 1;
  if (k <= Mid) Add(Index << 1, L, Mid, k);
  if (k > Mid) Add(Index << 1 | 1, Mid + 1, R, k);
  return;
}
int Query(int Index, int L, int R, int l, int r) {
  if (l <= L && R <= r) return Tree[Index];
  int Mid = L + R >> 1;
  if (r <= Mid) return Query(Index << 1, L, Mid, l, r);
  if (l > Mid) return Query(Index << 1 | 1, Mid + 1, R, l, r);
  return min(Query(Index << 1, L, Mid, l, r),
             Query(Index << 1 | 1, Mid + 1, R, l, r));
}
}  // namespace segmentTree
int Rec[Maxn];
struct Cmp {
  bool operator()(int x, int y) { return x > y; }
};
priority_queue<int, vector<int>, Cmp> PQ;
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) P[A[i]] = i;
  segmentTree::Build(1, 1, N);
  for (int i = 1; i <= N; ++i) {
    int x = segmentTree::Query(1, 1, N, P[i], min(N, P[i] + K - 1));
    segmentTree::Add(1, 1, N, P[i]);
    if (x == INF) continue;
    AddEdge(x, P[i]);
  }
  segmentTree::Build(1, 1, N);
  for (int i = N; i >= 1; --i) {
    int x = segmentTree::Query(1, 1, N, max(1, P[i] - K + 1), P[i]);
    segmentTree::Add(1, 1, N, P[i]);
    if (x == INF) continue;
    AddEdge(P[i], x);
  }
  for (int i = 1; i <= N; ++i)
    if (!Degree[i]) PQ.push(i);
  for (int i = 1; i <= N; ++i) {
    Rec[i] = PQ.top();
    PQ.pop();
    for (int t = Start[Rec[i]]; t; t = Edge[t].Next) {
      int v = Edge[t].To;
      --Degree[v];
      if (!Degree[v]) PQ.push(v);
    }
  }
  memset(A, 0, sizeof(A));
  for (int i = 1; i <= N; ++i) A[Rec[i]] = i;
  for (int i = 1; i <= N; ++i) printf("%d\n", A[i]);
  return 0;
}
