This problem can be solved with a data structure that supports the following operations:

• Add / Remove a value \(X\) as an element.
• Find the smallest element greater than a value \(X\).

For example, C++ supports a data structure called set for this purpose.

This time, we will explain an algorithm while maintaining the following information:
\(under_i\) … \(i\) the (face-up) card stacked right under \(i\) \(pile_i\) … the number of cards in the pile when \(i\) is stacked on the top of the pile

Process the cards in order by the following procedure.

• We will store the face-up topmost cards on the table in the data structure.
• Find the smallest face-up topmost card greater than \(X\) on the table (from the data structure.)
• If there is such a card, do the following.
  • Let \(under_X=Y\).
  • Let \(pile_X=pile_Y+1\).
  • Remove \(Y\) from the data structure and insert \(X\).
• If there is no such card, do the following, which simulates placing Card \(X\) without stacking onto anything:
  • Let \(pile_X=1\).
  • (You may set anything to \(under_X\).)
  • Insert \(X\) to the data structure.
• If \(pile_X=K\), then the pile has \(K\) cards, so eat them as follows:
  • First, we are going to eat the pile containing Card \(X\), so remove \(X\) from the data structure.
  • Then, eat Card \(X\). Now, Card \(under_X\) shows up face up, so let \(X \leftarrow under_X\). Repeat this until eating \(K\) cards.
• Repeat the steps above until running out of cards in the deck.

Sample code (C++):

#include<bits/stdc++.h>

using namespace std;

int main(){
  int n,k;
  cin >> n >> k;
  vector<int> under(n+5,-1);
  vector<int> pile(n+5,0);
  set<int> st;
  vector<int> res(n+5,-1);
  for(int i=1;i<=n;i++){
    int p;
    cin >> p;
    auto it=st.upper_bound(p);
    if(it==st.end()){
      pile[p]=1;
      st.insert(p);
    }
    else{
      under[p]=(*it);
      pile[p]=pile[(*it)]+1;
      st.erase(it);
      st.insert(p);
    }
    
    if(pile[p]==k){
      st.erase(p);
      int x=p;
      for(int j=0;j<k;j++){
        res[x]=i;
        x=under[x];
      }
    }
  }

  for(int i=1;i<=n;i++){cout << res[i] << "\n";}
  return 0;
}