#include <bits/stdc++.h>
using namespace std;
const int N = 205;
const int inf = 1e9;
int n, k, sum[N], t[N];
pair<int, pair<int, int> > p[N];
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].first >> p[i].second.first;
    p[i].second.second = i + 1;
    sum[p[i].first]++;
    sum[p[i].second.first + 1]--;
  }
  for (int i = 1; i < N; ++i) {
    sum[i] += sum[i - 1];
  }
  sort(p, p + n);
  for (int i = 1; i < N; ++i) {
    while (sum[i] > k) {
      int lst = -1, lstEnd = -1;
      for (int j = 0; j < n; ++j) {
        if (p[j].first > i) break;
        if (p[j].second.first < i) continue;
        if (t[j]) continue;
        if (p[j].second.first > lstEnd) lstEnd = p[j].second.first, lst = j;
      }
      t[lst] = 1;
      ans.push_back(p[lst].second.second);
      for (int j = p[lst].first; j <= p[lst].second.first; ++j) {
        sum[j]--;
      }
    }
  }
  cout << ans.size() << endl;
  for (auto cur : ans) cout << cur << ' ';
  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>
using namespace std;
const int N = 205;
const int inf = 1e9;
int n, k, sum[N], t[N];
pair<int, pair<int, int> > p[N];
vector<int> ans;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].first >> p[i].second.first;
    p[i].second.second = i + 1;
    sum[p[i].first]++;
    sum[p[i].second.first + 1]--;
  }
  for (int i = 1; i < N; ++i) {
    sum[i] += sum[i - 1];
  }
  sort(p, p + n);
  for (int i = 1; i < N; ++i) {
    while (sum[i] > k) {
      int lst = -1, lstEnd = -1;
      for (int j = 0; j < n; ++j) {
        if (p[j].first > i) break;
        if (p[j].second.first < i) continue;
        if (t[j]) continue;
        if (p[j].second.first > lstEnd) lstEnd = p[j].second.first, lst = j;
      }
      t[lst] = 1;
      ans.push_back(p[lst].second.second);
      for (int j = p[lst].first; j <= p[lst].second.first; ++j) {
        sum[j]--;
      }
    }
  }
  cout << ans.size() << endl;
  for (auto cur : ans) cout << cur << ' ';
  return 0;
}

```