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)

#include <bits/stdc++.h>

#ifdef ONLINE_JUDGE
#define endl "\n"
#endif
using namespace std;
typedef unsigned long long int lu;
typedef long long int ll;
typedef long double ld;
typedef vector<ll> v;
typedef vector<v> vv;
typedef vector<ld> vld;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
const ll MOD = 1e9 + 7;
const v dx = {1, -1, 0, 0};
const v dy = {0, 0, 1, -1};
#define yes cout << "YES\n";
#define no cout << "NO\n";
#define yesno(f) \
    if (f)       \
        yes else no;
#define noo \
    { no return; }
#define amax(x, y) \
    if (y > x)     \
        x = y;
#define amin(x, y) \
    if (y < x)     \
        x = y;
#define sz(x) (ll) x.size()
#define sq(x) ((x) * (x))

#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

template<typename T = ll>
vector<T> ga(ll n, bool oneIndexed = false) {
    vector<T> a = vector<T>(n + oneIndexed);

    for (ll i = 0; i < n; i++) {
        T p;
        cin >> p;
        a[i + oneIndexed] = p;
    }

    return move(a);
}

template<typename T, typename A>
void pa(vector<T, A> const &a, ll begin = 0, ll end = -1) {
    if (end == -1)
        end = sz(a) - 1;
    for (ll i = begin; i <= end; i++) {
        cout << a[i] << " ";
    }
    cout << endl;
}

v os;
ll C;

void solve() {
    ll n, k;
    cin >> n >> k >> C;
    string s;
    cin >> s;

    v posmap(n, 0);
    for (ll i = 0; i < n; i++) {
        if (s[i] == 'o') {
            posmap[i] = sz(os);
            os.push_back(i);
        }
    }
    if (k == 1) {
        if (sz(os) == 1) {
            cout << os[0];
        }
        return;
    }

    os.push_back(1e15);
    v dp(sz(os), 0);
    ll right = sz(os) - 1;

    for (ll day = sz(os) - 2; day >= 0; day--) {
        while (right > day && os[right] - os[day] > C) {
            right--;
        }
        right++;

        dp[day] = 1 + dp[right];
    }

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

    ll curr = 0;
    v res;
    ll cnt = 0;

    while (cnt < k) {
        while (s[curr] == 'x')curr++;

        if (dp[posmap[curr]] > dp[posmap[curr] + 1]) {
            res.push_back(curr);
        }

        curr += C + 1;
        cnt++;
    }
    for (auto x: res)cout << x + 1 << "\n";
}

int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    // ll t;cin >> t;while(t--)
    solve();
    return 0;
}