#include <bits/stdc++.h>
using namespace std;
int mycmp(const pair<pair<int, int>, int>& a,
          const pair<pair<int, int>, int>& b) {
  return a.first.first < a.first.first;
}
int main() {
  {
    cin.sync_with_stdio(0);
    cout.sync_with_stdio(0);
    cin.tie(0);
    cout.tie();
  }
  {}
  int n, k;
  cin >> n >> k;
  vector<int> f(300, 0);
  vector<pair<pair<int, int>, int> > a(n);
  for (int i = (int)(0); i < (int)(n); i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
    for (int j = (int)(a[i].first.first); j < (int)(a[i].first.second + 1); j++)
      f[j]++;
  }
  sort(a.begin(), a.end());
  vector<int> res(0);
  vector<int> check(n, 0);
  for (int i = 1; i <= 200; i++) {
    while (f[i] > k) {
      int Max = 0, dx = -1;
      for (int j = 0; j < n; j++) {
        if (check[j]) continue;
        if (a[j].first.first <= i && i <= a[j].first.second) {
          if (Max <= a[j].first.second - i) {
            Max = a[j].first.second - i;
            dx = j;
          }
        }
      }
      check[dx] = 1;
      res.push_back(a[dx].second);
      for (int j = (int)(a[dx].first.first); j < (int)(a[dx].first.second + 1);
           j++)
        f[j]--;
    }
  }
  sort(res.begin(), res.end());
  cout << res.size() << "\n";
  for (int i : res) cout << i << " ";
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
int mycmp(const pair<pair<int, int>, int>& a,
          const pair<pair<int, int>, int>& b) {
  return a.first.first < a.first.first;
}
int main() {
  {
    cin.sync_with_stdio(0);
    cout.sync_with_stdio(0);
    cin.tie(0);
    cout.tie();
  }
  {}
  int n, k;
  cin >> n >> k;
  vector<int> f(300, 0);
  vector<pair<pair<int, int>, int> > a(n);
  for (int i = (int)(0); i < (int)(n); i++) {
    cin >> a[i].first.first >> a[i].first.second;
    a[i].second = i + 1;
    for (int j = (int)(a[i].first.first); j < (int)(a[i].first.second + 1); j++)
      f[j]++;
  }
  sort(a.begin(), a.end());
  vector<int> res(0);
  vector<int> check(n, 0);
  for (int i = 1; i <= 200; i++) {
    while (f[i] > k) {
      int Max = 0, dx = -1;
      for (int j = 0; j < n; j++) {
        if (check[j]) continue;
        if (a[j].first.first <= i && i <= a[j].first.second) {
          if (Max <= a[j].first.second - i) {
            Max = a[j].first.second - i;
            dx = j;
          }
        }
      }
      check[dx] = 1;
      res.push_back(a[dx].second);
      for (int j = (int)(a[dx].first.first); j < (int)(a[dx].first.second + 1);
           j++)
        f[j]--;
    }
  }
  sort(res.begin(), res.end());
  cout << res.size() << "\n";
  for (int i : res) cout << i << " ";
  return 0;
}

```