#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int> > seg(n);
  vector<pair<pair<int, int>, int> > point;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    seg[i].first = a;
    seg[i].second = b;
    point.push_back({{a, 0}, i});
    point.push_back({{b, 1}, i});
  }
  sort(point.begin(), point.end());
  set<pair<int, int> > s;
  s.insert({seg[point[0].second].second, point[0].second});
  vector<int> ans;
  for (int i = 1; i < point.size(); i++) {
    if (s.size() > k) {
      auto it = s.end();
      it--;
      pair<int, int> temp = *it;
      ans.push_back(temp.second);
      s.erase(it);
    }
    if (point[i].first.second == 0) {
      s.insert({seg[point[i].second].second, point[i].second});
    } else {
      s.erase({point[i].first.first, point[i].second});
    }
  }
  if (s.size() > k) {
    auto it = s.end();
    it--;
    pair<int, int> temp = *it;
    ans.push_back(temp.second);
    s.erase(it);
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
}

### Prompt

In CPP, your task is to solve 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;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int> > seg(n);
  vector<pair<pair<int, int>, int> > point;
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    seg[i].first = a;
    seg[i].second = b;
    point.push_back({{a, 0}, i});
    point.push_back({{b, 1}, i});
  }
  sort(point.begin(), point.end());
  set<pair<int, int> > s;
  s.insert({seg[point[0].second].second, point[0].second});
  vector<int> ans;
  for (int i = 1; i < point.size(); i++) {
    if (s.size() > k) {
      auto it = s.end();
      it--;
      pair<int, int> temp = *it;
      ans.push_back(temp.second);
      s.erase(it);
    }
    if (point[i].first.second == 0) {
      s.insert({seg[point[i].second].second, point[i].second});
    } else {
      s.erase({point[i].first.first, point[i].second});
    }
  }
  if (s.size() > k) {
    auto it = s.end();
    it--;
    pair<int, int> temp = *it;
    ans.push_back(temp.second);
    s.erase(it);
  }
  sort(ans.begin(), ans.end());
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
}

```