Problem:
Score : 500 points

Problem Statement

Takahashi has decided to work on K days of his choice from the N days starting with tomorrow.

You are given an integer C and a string S. Takahashi will choose his workdays as follows:

After working for a day, he will refrain from working on the subsequent C days.

If the i-th character of S is x, he will not work on Day i, where Day 1 is tomorrow, Day 2 is the day after tomorrow, and so on.

Find all days on which Takahashi is bound to work.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq N

0 \leq C \leq N

The length of S is N.

Each character of S is o or x.

Takahashi can choose his workdays so that the conditions in Problem Statement are satisfied.

Input

Input is given from Standard Input in the following format:

N K C
S

Output

Print all days on which Takahashi is bound to work in ascending order, one per line.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h> 
using namespace std; 
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define flash ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define debug(x) cerr << " - " << #x << ": " << x << endl;
#define debugs(x, y) cerr << " - " << #x << ": " << x << " " << #y << ": " << y << endl;
#define all(x) (x).begin(),(x).end()
#define sz(x) (ll)x.size()
#define ll long long
#define INF 1000000000
#define MOD 1000000007
#define pb push_back
#define ve vector<ll>
#define dos pair<ll,ll>
#define vedos vector<dos>
#define rand mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define EPS 0.000001
struct greateri
{
    template<class T>
    bool operator()(T const &a, T const &b) const { return a > b; }
};
void setIO(string s) {
    ios_base::sync_with_stdio(0); cin.tie(0); 
    freopen((s+".in").c_str(),"r",stdin);
    freopen((s+".out").c_str(),"w",stdout);
}
int dp[300001],dp1[300001];
string kal;
int c,n,k;
set<int>pos;
set<int>posi;
int solve(int index)
{
    int ans=1;
    auto u = pos.lower_bound(index+c+1);
    if(dp[index]!=-1)return dp[index];
    if(u!=pos.end())
    {
       ans=max(ans,1+solve((*u))); 
    }
    return dp[index]=ans;
}
int solve1(int index)
{
    int ans=1;
    auto u = posi.lower_bound(-index+c+1);
    if(dp1[index]!=-1)return dp1[index];
    if(u!=pos.end() && (*u)<=0)
    {
       ans=max(ans,1+solve1(-(*u))); 
    }
    return dp1[index]=ans;
}
int main()
{
  //flash;
  cin>>n>>k>>c;
  cin>>kal;
  ve kaki;
  memset(dp,-1,sizeof dp);
  memset(dp1,-1,sizeof dp1);
  for (int i = 0; i < kal.size(); ++i)
  {
      if(kal[i]=='o'){pos.insert(i);posi.insert(-i);kaki.pb(i);}
  }
  for(auto it:pos){solve(it);solve1(it);}
  int last=0;
  ve ans;
  for (int i = 0; i < kaki.size(); ++i)
  {
    int v=0;
    int s=0;
    if((i+1)>=kaki.size())s=0;
    else s = dp[kaki[i+1]];
    if((i-1)<0)v=0;
    else v = dp1[kaki[i-1]];
    if((v+s)<k)ans.pb(kaki[i]);
  }
  for(auto it:ans)cout<<it+1<<'\n';
  return 0;
}
//code the AC sol !
// BS/queue/map