#include <bits/stdc++.h>
using namespace std;
struct interval {
  int left;
  int right;
  int id;
};
bool mysort(interval a, interval b) {
  if (a.left == b.left)
    return a.right > b.right;
  else
    return a.left < b.left;
}
struct comp {
  bool operator()(interval a, interval b) { return a.right < b.right; }
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<interval> vect;
  int pre[202] = {0};
  int ppre[202] = {0};
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    pre[x]++;
    pre[y + 1]--;
    struct interval temp;
    temp.left = x;
    temp.right = y;
    temp.id = i + 1;
    vect.push_back(temp);
  }
  for (int i = 1; i < 201; i++) pre[i] += pre[i - 1];
  sort(vect.begin(), vect.end(), mysort);
  int count = 0;
  vector<int> ans;
  int index = 0;
  priority_queue<interval, vector<interval>, comp> pq;
  for (int i = 0; i < 201; i++) {
    if (i > 0) {
      ppre[i] += ppre[i - 1];
      pre[i] += ppre[i];
    }
    if (pre[i] > k) {
      for (int j = index; j < n; j++) {
        if (vect[j].left <= i && vect[j].right >= i) {
          pq.push(vect[j]);
          index++;
        } else if (vect[j].left > i)
          break;
        else if (vect[j].right < i)
          index++;
      }
      while (pre[i] > k && !pq.empty()) {
        ppre[i]--;
        pre[i]--;
        count++;
        ans.push_back(pq.top().id);
        ppre[pq.top().right + 1]++;
        pq.pop();
      }
    }
  }
  cout << count << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  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;
struct interval {
  int left;
  int right;
  int id;
};
bool mysort(interval a, interval b) {
  if (a.left == b.left)
    return a.right > b.right;
  else
    return a.left < b.left;
}
struct comp {
  bool operator()(interval a, interval b) { return a.right < b.right; }
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<interval> vect;
  int pre[202] = {0};
  int ppre[202] = {0};
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    pre[x]++;
    pre[y + 1]--;
    struct interval temp;
    temp.left = x;
    temp.right = y;
    temp.id = i + 1;
    vect.push_back(temp);
  }
  for (int i = 1; i < 201; i++) pre[i] += pre[i - 1];
  sort(vect.begin(), vect.end(), mysort);
  int count = 0;
  vector<int> ans;
  int index = 0;
  priority_queue<interval, vector<interval>, comp> pq;
  for (int i = 0; i < 201; i++) {
    if (i > 0) {
      ppre[i] += ppre[i - 1];
      pre[i] += ppre[i];
    }
    if (pre[i] > k) {
      for (int j = index; j < n; j++) {
        if (vect[j].left <= i && vect[j].right >= i) {
          pq.push(vect[j]);
          index++;
        } else if (vect[j].left > i)
          break;
        else if (vect[j].right < i)
          index++;
      }
      while (pre[i] > k && !pq.empty()) {
        ppre[i]--;
        pre[i]--;
        count++;
        ans.push_back(pq.top().id);
        ppre[pq.top().right + 1]++;
        pq.pop();
      }
    }
  }
  cout << count << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}

```