#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5;
bool segcomp(pair<pair<int, int>, int> s1, pair<pair<int, int>, int> s2) {
  if (s1.first.second == s2.first.second)
    return s1.first.first < s2.first.first;
  return s1.first.second > s2.first.second;
}
int p[maxn + 1];
vector<pair<pair<int, int>, int> > segments[maxn + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  pair<pair<int, int>, int> temp;
  int firstSegL = 0;
  for (int i = 0; i < n; i++) {
    cin >> temp.first.first >> temp.first.second;
    if (firstSegL == 0) firstSegL = temp.first.first;
    temp.second = i + 1;
    for (int j = temp.first.first; j <= temp.first.second; j++)
      segments[j].push_back(temp);
  }
  for (int i = 1; i <= maxn; i++)
    sort(segments[i].begin(), segments[i].end(), segcomp);
  int res = 0;
  set<int> res_idx;
  for (int i = 1; i <= maxn; i++)
    while (segments[i].size() > k) {
      res++;
      pair<pair<int, int>, int> temp = segments[i].front();
      res_idx.insert(temp.second);
      for (int j = temp.first.first; j <= temp.first.second; j++)
        for (int k = 0; k < segments[j].size(); k++)
          if (segments[j][k].first.first == temp.first.first &&
              segments[j][k].first.second == temp.first.second &&
              segments[j][k].second == temp.second) {
            segments[j].erase(segments[j].begin() + k);
            break;
          }
    }
  cout << res << "\n";
  for (auto out : res_idx) cout << out << " ";
  cout << "\n";
}

### Prompt

Your task is to create 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;
const int maxn = 2e5;
bool segcomp(pair<pair<int, int>, int> s1, pair<pair<int, int>, int> s2) {
  if (s1.first.second == s2.first.second)
    return s1.first.first < s2.first.first;
  return s1.first.second > s2.first.second;
}
int p[maxn + 1];
vector<pair<pair<int, int>, int> > segments[maxn + 1];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  pair<pair<int, int>, int> temp;
  int firstSegL = 0;
  for (int i = 0; i < n; i++) {
    cin >> temp.first.first >> temp.first.second;
    if (firstSegL == 0) firstSegL = temp.first.first;
    temp.second = i + 1;
    for (int j = temp.first.first; j <= temp.first.second; j++)
      segments[j].push_back(temp);
  }
  for (int i = 1; i <= maxn; i++)
    sort(segments[i].begin(), segments[i].end(), segcomp);
  int res = 0;
  set<int> res_idx;
  for (int i = 1; i <= maxn; i++)
    while (segments[i].size() > k) {
      res++;
      pair<pair<int, int>, int> temp = segments[i].front();
      res_idx.insert(temp.second);
      for (int j = temp.first.first; j <= temp.first.second; j++)
        for (int k = 0; k < segments[j].size(); k++)
          if (segments[j][k].first.first == temp.first.first &&
              segments[j][k].first.second == temp.first.second &&
              segments[j][k].second == temp.second) {
            segments[j].erase(segments[j].begin() + k);
            break;
          }
    }
  cout << res << "\n";
  for (auto out : res_idx) cout << out << " ";
  cout << "\n";
}

```