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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define ll long long
#define pb push_back
#define PII pair<int,int>
#define X first
#define Y second
#define MP make_pair
#define PI acos(-1.0)
#define eps 1e-8
const int maxn=3e5+10;
const int mod=998244353;
using namespace std;

int n,k,c;
int dp[maxn],vis[maxn],start[maxn],end[maxn];
char str[maxn];
stack<int>s;
vector<int>g;
set<int>ans;

int main()
{
    scanf("%d%d%d%s",&n,&k,&c,str+1);
    int num=0;
    for(int i=1;i<=n;i++)
    {
        if(i<=c+1)
            dp[i]=1;
        else 
            if(str[i]=='x')
                dp[i]=dp[i-1];
            else 
                dp[i]=max(dp[i-1],dp[i-c-1]+1);
        // if(dp[i]>dp[i-1])
        //     end[dp[i-1]]=i-1;
        if(str[i]=='o')
        {
            if(vis[dp[i]]==0)
                start[dp[i]]=i;
            vis[dp[i]]++;
            g.pb(i);
        }
        // printf("%d %d\n",i,dp[i]);
    }
    if(dp[n]!=k)
        return 0;
    for(int i=1;i<=n;i++)
    {
        if(str[i]=='o'&&vis[dp[i]]==1)
        {
            ans.insert(i);
            int tmp=i;
            while(!ans.count(start[dp[tmp]-1])&&dp[tmp]>1&&upper_bound(g.begin(),g.end(),tmp-c-1)-upper_bound(g.begin(),g.end(),start[dp[tmp]-1])==0)
                ans.insert(start[dp[tmp]-1]),tmp=start[dp[tmp]-1];
            // tmp=i;
            // while(dp[tmp]<k&&upper_bound(g.begin(),g.end(),tmp+c+1)-lower_bound(g.begin(),g.end(),tmp+c+1)==0)
                // ans.insert(start[dp[tmp]+1]),tmp=start[dp[tmp]+1];
        }
            
        
        
            
        

    }
    for(set<int>::iterator it=ans.begin();it!=ans.end();it++)
        printf("%d\n",*it);
    // system("pause");
    return 0;
}