#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7, siz = 1e6 + 5;
long long t, n, m, k, en[siz], a[siz];
pair<long long, long long> p[siz];
bool gone[siz];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    en[i + 1] = r;
    p[2 * i].first = l;
    p[2 * i + 1].first = r;
    p[2 * i].second = -(i + 1);
    p[2 * i + 1].second = i + 1;
  }
  sort(p, p + 2 * n);
  long long have = 0;
  vector<long long> del;
  set<pair<long long, long long>, greater<pair<long long, long long> > > s;
  for (long long i = 0; i < 2 * n; i++) {
    long long ind = abs(p[i].second);
    if (p[i].second > 0) {
      if (!gone[ind]) {
        have--;
        s.erase({en[ind], ind});
      }
    } else {
      have++;
      s.insert({en[ind], ind});
      if (have > k) {
        pair<long long, long long> p = *s.begin();
        s.erase(s.begin());
        gone[p.second] = 1;
        del.push_back(p.second);
        have--;
      }
    }
  }
  cout << del.size() << endl;
  for (auto x : del) cout << x << " ";
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 200) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 200) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
1 4 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 2 4 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7, siz = 1e6 + 5;
long long t, n, m, k, en[siz], a[siz];
pair<long long, long long> p[siz];
bool gone[siz];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    en[i + 1] = r;
    p[2 * i].first = l;
    p[2 * i + 1].first = r;
    p[2 * i].second = -(i + 1);
    p[2 * i + 1].second = i + 1;
  }
  sort(p, p + 2 * n);
  long long have = 0;
  vector<long long> del;
  set<pair<long long, long long>, greater<pair<long long, long long> > > s;
  for (long long i = 0; i < 2 * n; i++) {
    long long ind = abs(p[i].second);
    if (p[i].second > 0) {
      if (!gone[ind]) {
        have--;
        s.erase({en[ind], ind});
      }
    } else {
      have++;
      s.insert({en[ind], ind});
      if (have > k) {
        pair<long long, long long> p = *s.begin();
        s.erase(s.begin());
        gone[p.second] = 1;
        del.push_back(p.second);
        have--;
      }
    }
  }
  cout << del.size() << endl;
  for (auto x : del) cout << x << " ";
}

```