#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream& operator<<(ostream& Stream, const vector<T>& Vec) {
  for (const T& i : Vec) Stream << i << " ";
  return Stream;
}
template <typename A, typename B>
ostream& operator<<(ostream& Stream, const pair<A, B>& Pair) {
  return Stream << Pair.first << ' ' << Pair.second;
}
template <typename A, typename B, typename C>
ostream& operator<<(ostream& Stream, const tuple<A, B, C>& T) {
  return Stream << get<0>(T) << ' ' << get<1>(T) << ' ' << get<2>(T);
}
const int N = 200002;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<pair<int, int>>> begin(N);
  vector<vector<int>> end(N);
  vector<bool> vis(N, false);
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    begin[l].push_back({r, i + 1});
    end[r].push_back(i + 1);
  }
  priority_queue<pair<int, int>> pq;
  set<int> cur;
  vector<int> answer;
  for (int i = 1; i < N; ++i) {
    for (auto& p : begin[i]) {
      pq.push(p);
      cur.insert(p.second);
    }
    while (int(cur.size()) > k) {
      auto p = pq.top();
      pq.pop();
      if (vis[p.second]) continue;
      if (cur.find(p.second) != cur.end()) {
        cur.erase(p.second);
        answer.emplace_back(p.second);
      }
    }
    for (auto& x : end[i]) {
      vis[x] = true;
      if (cur.find(x) != cur.end()) cur.erase(x);
    }
  }
  cout << answer.size() << '\n' << answer;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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;
template <typename T>
ostream& operator<<(ostream& Stream, const vector<T>& Vec) {
  for (const T& i : Vec) Stream << i << " ";
  return Stream;
}
template <typename A, typename B>
ostream& operator<<(ostream& Stream, const pair<A, B>& Pair) {
  return Stream << Pair.first << ' ' << Pair.second;
}
template <typename A, typename B, typename C>
ostream& operator<<(ostream& Stream, const tuple<A, B, C>& T) {
  return Stream << get<0>(T) << ' ' << get<1>(T) << ' ' << get<2>(T);
}
const int N = 200002;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<vector<pair<int, int>>> begin(N);
  vector<vector<int>> end(N);
  vector<bool> vis(N, false);
  for (int i = 0; i < n; ++i) {
    int l, r;
    cin >> l >> r;
    begin[l].push_back({r, i + 1});
    end[r].push_back(i + 1);
  }
  priority_queue<pair<int, int>> pq;
  set<int> cur;
  vector<int> answer;
  for (int i = 1; i < N; ++i) {
    for (auto& p : begin[i]) {
      pq.push(p);
      cur.insert(p.second);
    }
    while (int(cur.size()) > k) {
      auto p = pq.top();
      pq.pop();
      if (vis[p.second]) continue;
      if (cur.find(p.second) != cur.end()) {
        cur.erase(p.second);
        answer.emplace_back(p.second);
      }
    }
    for (auto& x : end[i]) {
      vis[x] = true;
      if (cur.find(x) != cur.end()) cur.erase(x);
    }
  }
  cout << answer.size() << '\n' << answer;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout.precision(10);
  cout << fixed;
  int t = 1;
  while (t--) solve();
  return 0;
}

```