#include <bits/stdc++.h>
using namespace std;
signed main() {
  std ::ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long mxn = 200050;
  long long n, k;
  cin >> n >> k;
  vector<long long> tot(mxn);
  vector<pair<long long, long long>> lr(n + 1);
  vector<vector<long long>> nxt(mxn);
  for (long long i = 1; i <= n; i++) {
    cin >> lr[i].first >> lr[i].second;
    tot[lr[i].first]++;
    tot[lr[i].second + 1]--;
    nxt[lr[i].first].push_back(i);
    nxt[lr[i].second + 1].push_back(-i);
  }
  for (long long i = 1; i < mxn; i++) tot[i] += tot[i - 1];
  set<pair<long long, long long>> tmp;
  vector<long long> rmv(mxn);
  vector<bool> res(n + 1);
  long long del = 0;
  long long ans = 0;
  for (long long i = 1; i < mxn; i++) {
    del += rmv[i];
    for (long long x : nxt[i]) {
      if (x > 0)
        tmp.insert({lr[x].second, x});
      else {
        auto ff = tmp.find({lr[-x].second, -x});
        if (ff != tmp.end()) tmp.erase(ff);
      }
    }
    while (tot[i] - del > k) {
      auto t = tmp.rbegin();
      long long pos = t->second;
      tmp.erase(prev(tmp.end()));
      rmv[lr[pos].second + 1]--;
      del++;
      res[pos] = true;
      ans++;
    }
  }
  cout << ans << "\n";
  for (long long i = 1; i <= n; i++)
    if (res[i]) cout << i << " ";
  cout << "\n";
}

### Prompt

Generate a cpp solution to 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;
signed main() {
  std ::ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long mxn = 200050;
  long long n, k;
  cin >> n >> k;
  vector<long long> tot(mxn);
  vector<pair<long long, long long>> lr(n + 1);
  vector<vector<long long>> nxt(mxn);
  for (long long i = 1; i <= n; i++) {
    cin >> lr[i].first >> lr[i].second;
    tot[lr[i].first]++;
    tot[lr[i].second + 1]--;
    nxt[lr[i].first].push_back(i);
    nxt[lr[i].second + 1].push_back(-i);
  }
  for (long long i = 1; i < mxn; i++) tot[i] += tot[i - 1];
  set<pair<long long, long long>> tmp;
  vector<long long> rmv(mxn);
  vector<bool> res(n + 1);
  long long del = 0;
  long long ans = 0;
  for (long long i = 1; i < mxn; i++) {
    del += rmv[i];
    for (long long x : nxt[i]) {
      if (x > 0)
        tmp.insert({lr[x].second, x});
      else {
        auto ff = tmp.find({lr[-x].second, -x});
        if (ff != tmp.end()) tmp.erase(ff);
      }
    }
    while (tot[i] - del > k) {
      auto t = tmp.rbegin();
      long long pos = t->second;
      tmp.erase(prev(tmp.end()));
      rmv[lr[pos].second + 1]--;
      del++;
      res[pos] = true;
      ans++;
    }
  }
  cout << ans << "\n";
  for (long long i = 1; i <= n; i++)
    if (res[i]) cout << i << " ";
  cout << "\n";
}

```