#include <bits/stdc++.h>
using namespace std;
const int N = 200200;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> segs(n);
  vector<int> cnt(N);
  vector<vector<int>> ev(N);
  for (int i = 0; i < n; ++i) {
    cin >> segs[i].first >> segs[i].second;
    ++cnt[segs[i].first];
    --cnt[segs[i].second + 1];
    ev[segs[i].first].push_back(i + 1);
    ev[segs[i].second + 1].push_back(-i - 1);
  }
  for (int i = 0; i + 1 < N; ++i) {
    cnt[i + 1] += cnt[i];
  }
  vector<int> ans(n), sub(N);
  set<pair<int, int>> curSegs;
  int curSub = 0;
  for (int i = 0; i < N; ++i) {
    curSub += sub[i];
    for (auto it : ev[i]) {
      if (it > 0) {
        curSegs.insert(make_pair(segs[it - 1].second, it - 1));
      } else {
        auto iter = curSegs.find(make_pair(segs[-it - 1].second, -it - 1));
        if (iter != curSegs.end()) curSegs.erase(iter);
      }
    }
    while (cnt[i] - curSub > k) {
      assert(!curSegs.empty());
      int pos = curSegs.rbegin()->second;
      curSegs.erase(prev(curSegs.end()));
      ++curSub;
      --sub[segs[pos].second + 1];
      ans[pos] = 1;
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << endl;
  for (int i = 0; i < n; ++i) {
    if (ans[i]) cout << i + 1 << " ";
  }
  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;
const int N = 200200;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> segs(n);
  vector<int> cnt(N);
  vector<vector<int>> ev(N);
  for (int i = 0; i < n; ++i) {
    cin >> segs[i].first >> segs[i].second;
    ++cnt[segs[i].first];
    --cnt[segs[i].second + 1];
    ev[segs[i].first].push_back(i + 1);
    ev[segs[i].second + 1].push_back(-i - 1);
  }
  for (int i = 0; i + 1 < N; ++i) {
    cnt[i + 1] += cnt[i];
  }
  vector<int> ans(n), sub(N);
  set<pair<int, int>> curSegs;
  int curSub = 0;
  for (int i = 0; i < N; ++i) {
    curSub += sub[i];
    for (auto it : ev[i]) {
      if (it > 0) {
        curSegs.insert(make_pair(segs[it - 1].second, it - 1));
      } else {
        auto iter = curSegs.find(make_pair(segs[-it - 1].second, -it - 1));
        if (iter != curSegs.end()) curSegs.erase(iter);
      }
    }
    while (cnt[i] - curSub > k) {
      assert(!curSegs.empty());
      int pos = curSegs.rbegin()->second;
      curSegs.erase(prev(curSegs.end()));
      ++curSub;
      --sub[segs[pos].second + 1];
      ans[pos] = 1;
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << endl;
  for (int i = 0; i < n; ++i) {
    if (ans[i]) cout << i + 1 << " ";
  }
  cout << endl;
  return 0;
}

```