#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, unsigned long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int powermod(long long int x, unsigned long long int y,
                       long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int32_t main() {
  long long int n, k;
  cin >> n >> k;
  vector<pair<long long int, long long int> > vp[200005];
  vector<long long int> ans;
  for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
    long long int l, r;
    cin >> l >> r;
    vp[l].push_back({r, i + 1});
  }
  set<pair<long long int, long long int> > s;
  for (long long int i = (long long int)(1); i < (long long int)(2e5 + 1);
       i++) {
    for (auto &val : vp[i]) {
      s.insert(val);
    }
    while (s.size() > k) {
      auto last = *prev(s.end());
      ans.push_back(last.second);
      s.erase(prev(s.end()));
    }
    while (s.size()) {
      auto f = *s.begin();
      if (f.first == i)
        s.erase(s.begin());
      else
        break;
    }
  }
  cout << ans.size() << "\n";
  for (const auto itr : ans) {
    cout << itr << ' ';
  }
  cout << "\n";
  ;
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long int power(long long int x, unsigned long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
long long int powermod(long long int x, unsigned long long int y,
                       long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) {
      res = (res * x) % p;
    }
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int32_t main() {
  long long int n, k;
  cin >> n >> k;
  vector<pair<long long int, long long int> > vp[200005];
  vector<long long int> ans;
  for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
    long long int l, r;
    cin >> l >> r;
    vp[l].push_back({r, i + 1});
  }
  set<pair<long long int, long long int> > s;
  for (long long int i = (long long int)(1); i < (long long int)(2e5 + 1);
       i++) {
    for (auto &val : vp[i]) {
      s.insert(val);
    }
    while (s.size() > k) {
      auto last = *prev(s.end());
      ans.push_back(last.second);
      s.erase(prev(s.end()));
    }
    while (s.size()) {
      auto f = *s.begin();
      if (f.first == i)
        s.erase(s.begin());
      else
        break;
    }
  }
  cout << ans.size() << "\n";
  for (const auto itr : ans) {
    cout << itr << ' ';
  }
  cout << "\n";
  ;
  return 0;
}

```