#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5;
int dist[N];
vector<int> open[N], close[N];
int n, k;
bool done[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (auto i = 0; i < (long long)(n); i++) {
    int l, r;
    cin >> l >> r;
    open[l].push_back(i);
    close[r + 1].push_back(i);
    dist[i] = r + 1;
  }
  set<int> pool;
  set<pair<int, int> > dist_idx;
  vector<int> res;
  for (auto i = 1; i < (long long)(N); i++) {
    for (auto j : close[i]) {
      if (done[j]) continue;
      pool.erase(j);
      dist_idx.erase({dist[j], j});
    }
    for (auto j : open[i]) {
      pool.insert(j);
      dist_idx.insert({dist[j], j});
    }
    while (((int)((pool).size())) > k) {
      auto it = dist_idx.end();
      it--;
      int idx = (*it).second;
      res.push_back(idx + 1);
      dist_idx.erase(it);
      pool.erase(idx);
      done[idx] = true;
    }
  }
  cout << ((int)((res).size())) << '\n';
  for (auto i : res) cout << i << " ";
  cout << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 2e5 + 5;
int dist[N];
vector<int> open[N], close[N];
int n, k;
bool done[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (auto i = 0; i < (long long)(n); i++) {
    int l, r;
    cin >> l >> r;
    open[l].push_back(i);
    close[r + 1].push_back(i);
    dist[i] = r + 1;
  }
  set<int> pool;
  set<pair<int, int> > dist_idx;
  vector<int> res;
  for (auto i = 1; i < (long long)(N); i++) {
    for (auto j : close[i]) {
      if (done[j]) continue;
      pool.erase(j);
      dist_idx.erase({dist[j], j});
    }
    for (auto j : open[i]) {
      pool.insert(j);
      dist_idx.insert({dist[j], j});
    }
    while (((int)((pool).size())) > k) {
      auto it = dist_idx.end();
      it--;
      int idx = (*it).second;
      res.push_back(idx + 1);
      dist_idx.erase(it);
      pool.erase(idx);
      done[idx] = true;
    }
  }
  cout << ((int)((res).size())) << '\n';
  for (auto i : res) cout << i << " ";
  cout << '\n';
  return 0;
}

```