#include <bits/stdc++.h>
using namespace std;
using pii = pair<long long, long long>;
ifstream in("in.in");
const long long INF = 1e9;
void speed_up() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
struct segm {
  long long b, e, id = 1;
};
bool operator<(const segm &a, const segm &b) {
  if (a.e != b.e) {
    return a.e > b.e;
  }
  if (a.b != b.b) {
    return a.b < b.b;
  }
  return a.id < b.id;
}
struct event {
  long long x, id, type;
};
bool operator<(const event &a, const event &b) {
  return (a.x == b.x) ? (a.type < b.type) : a.x < b.x;
}
signed main() {
  speed_up();
  long long n, k;
  cin >> n >> k;
  vector<segm> v(n);
  vector<event> events;
  events.reserve(2 * n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].b >> v[i].e;
    v[i].id = i;
    events.push_back({v[i].b, i, 0});
    events.push_back({v[i].e, i, 1});
  }
  sort(events.begin(), events.end());
  set<segm> s;
  set<long long> deleted;
  long long now = 0;
  for (event e : events) {
    if (deleted.count(e.id)) {
      continue;
    }
    if (e.type == 0) {
      if (now == k) {
        s.insert(v[e.id]);
        auto mini = *s.begin();
        s.erase(mini);
        deleted.insert(mini.id);
      } else {
        now++;
        s.insert(v[e.id]);
      }
    } else {
      now--;
      s.erase(v[e.id]);
    }
  }
  cout << deleted.size() << "\n";
  for (long long i : deleted) {
    cout << i + 1 << " ";
  }
  cout << "\n";
  return 0;
}

### 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;
using pii = pair<long long, long long>;
ifstream in("in.in");
const long long INF = 1e9;
void speed_up() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
struct segm {
  long long b, e, id = 1;
};
bool operator<(const segm &a, const segm &b) {
  if (a.e != b.e) {
    return a.e > b.e;
  }
  if (a.b != b.b) {
    return a.b < b.b;
  }
  return a.id < b.id;
}
struct event {
  long long x, id, type;
};
bool operator<(const event &a, const event &b) {
  return (a.x == b.x) ? (a.type < b.type) : a.x < b.x;
}
signed main() {
  speed_up();
  long long n, k;
  cin >> n >> k;
  vector<segm> v(n);
  vector<event> events;
  events.reserve(2 * n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].b >> v[i].e;
    v[i].id = i;
    events.push_back({v[i].b, i, 0});
    events.push_back({v[i].e, i, 1});
  }
  sort(events.begin(), events.end());
  set<segm> s;
  set<long long> deleted;
  long long now = 0;
  for (event e : events) {
    if (deleted.count(e.id)) {
      continue;
    }
    if (e.type == 0) {
      if (now == k) {
        s.insert(v[e.id]);
        auto mini = *s.begin();
        s.erase(mini);
        deleted.insert(mini.id);
      } else {
        now++;
        s.insert(v[e.id]);
      }
    } else {
      now--;
      s.erase(v[e.id]);
    }
  }
  cout << deleted.size() << "\n";
  for (long long i : deleted) {
    cout << i + 1 << " ";
  }
  cout << "\n";
  return 0;
}

```