#include <bits/stdc++.h>
using namespace std;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  set<tuple<int, bool, int>> queue;
  vector<pair<int, int>> planks(N);
  int a, b;
  for (int i = 0; i < N; ++i) {
    cin >> a >> b;
    planks[i] = {a, b};
    queue.insert({a, 0, i});
    queue.insert({b, 1, i});
  }
  set<int> remove;
  set<pair<int, int>> current;
  int loc, num;
  bool type;
  while (!queue.empty()) {
    tie(loc, type, num) = *queue.begin();
    queue.erase(queue.begin());
    if (type == 0) {
      current.insert({loc, num});
    } else if (current.find({planks[num].first, num}) != current.end()) {
      current.erase(current.find({planks[num].first, num}));
    }
    if (current.size() > K) {
      set<pair<int, int>>::iterator best = current.begin();
      for (auto it = current.begin(); it != current.end(); ++it) {
        if (planks[(*it).second].second > planks[(*best).second].second) {
          best = it;
        }
      }
      remove.insert((*best).second);
      current.erase(best);
    }
  }
  cout << remove.size() << '\n';
  for (auto ele : remove) {
    cout << ele + 1 << ' ';
  }
  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;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  set<tuple<int, bool, int>> queue;
  vector<pair<int, int>> planks(N);
  int a, b;
  for (int i = 0; i < N; ++i) {
    cin >> a >> b;
    planks[i] = {a, b};
    queue.insert({a, 0, i});
    queue.insert({b, 1, i});
  }
  set<int> remove;
  set<pair<int, int>> current;
  int loc, num;
  bool type;
  while (!queue.empty()) {
    tie(loc, type, num) = *queue.begin();
    queue.erase(queue.begin());
    if (type == 0) {
      current.insert({loc, num});
    } else if (current.find({planks[num].first, num}) != current.end()) {
      current.erase(current.find({planks[num].first, num}));
    }
    if (current.size() > K) {
      set<pair<int, int>>::iterator best = current.begin();
      for (auto it = current.begin(); it != current.end(); ++it) {
        if (planks[(*it).second].second > planks[(*best).second].second) {
          best = it;
        }
      }
      remove.insert((*best).second);
      current.erase(best);
    }
  }
  cout << remove.size() << '\n';
  for (auto ele : remove) {
    cout << ele + 1 << ' ';
  }
  return 0;
}

```