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: Wrong Answer)

/*                         _
                        _ooOoo_
                       o8888888o
                       88" . "88
                       (| -_- |)
                       O\  =  /O
                    ____/`---'\____
                  .'  \\|     |//  `.
                 /  \\|||  :  |||//  \
                /  _||||| -:- |||||_  \
                |   | \\\  -  /'| |   |
                | \_|  `\`---'//  |_/ |
                \  .-\__ `-. -'__/-.  /
              ___`. .'  /--.--\  `. .'___
           ."" '<  `.___\_<|>_/___.' _> \"".
          | | :  `- \`. ;`. _/; .'/ /  .' ; |
          \  \ `-.   \_\_`. _.'_/_/  -' _.' /
===========`-.`___`-.__\ \___  /__.-'_.'_.-'================

                  Please give me AC.
*/

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
#include <complex>
#include <bitset>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <chrono>
#include <random>

using namespace std;

using int64 = long long;
using uint64 = unsigned long long;
using vi = vector<int>;
using vl = vector<int64>;
using pii = pair<int, int>;
using pll = pair<int64, int64>;

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(v) (v).begin(), (v).end()
#define print(x) cout << (x) << '\n'
#define print2(x, y) cout << (x) << ' ' << (y) << '\n'
#define print3(x, y, z) cout << (x) << ' ' << (y) << ' ' << (z) << '\n'
#define printn(v) rep(i, (v).size() - 1) cout << (v)[i] << ' '; cout << (v)[n - 1] << '\n';
#define debug(x) cerr << #x << ": " << (x) << '\n'
#define debug2(x, y) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << '\n'
#define debug3(x, y, z) cerr << #x << ": " << (x) << ", " << #y << ": " << (y) << ", " << #z << ": " << (z) << '\n'
#define dbg(v) for (size_t _ = 0; _ < v.size(); ++_){cerr << #v << "[" << _ << "] : " << v[_] << '\n';}
// constant
const int INF = (1<<30) - 1;
const int64 INF64 = (1LL<<62) - 1;

template<typename T> T gcd(T a, T b) {
    if (a < b) return gcd(b, a);
    T r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

template<typename T> T lcm(const T a, const T b) {
    return a / gcd(a, b) * b;
}

template<typename T> bool chmin(T& a, const T& b) {
    if (a > b) return a = b, true; else return false;
}

template<typename T> bool chmax(T& a, const T& b) {
    if (a < b) return a = b, true; else return false;
}

// End of template.

int dp[400200];
int cup[200200];
vector<int> ans;

int main() {
    cout << fixed << setprecision(15);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    // cin.read(buf, sizeof buf); // 注意: ./a.out < in か pbp | ./a.out で入力すること

    int n, k, c;
    string s;
    cin >> n >> k >> c >> s;

    int t = 0;

    for (int i = n - 1; i >= 0; --i) {
        if (s[i] == 'o') dp[i] = t + 1; else dp[i] = -1;

        chmax(t, dp[i + c]);
    }

    // rep(i, n) cerr << dp[i] << ' '; cerr << '\n';

    if (dp[0] > k) {
        return 0;
    }

    for (int i = n - 1; i >= 0; --i) cup[i] = max(cup[i + 1], dp[i]);

    for (int i = 0; i < n;) {
        int x = dp[i];
        if (cup[i + 1] != dp[i]) {
            ans.push_back(i);
        }
        i += c + 1;
        while (i < n && dp[i] != x - 1) {++i;}
    }


    for (const auto x : ans) print(x + 1);

    return 0;
}