#include <bits/stdc++.h>
using namespace std;
using ll = long long;
struct Seg {
  int s, t, n;
  Seg(int s = 0, int t = 0, int n = 0) : s(s), t(t), n(n) {}
  bool operator<(const Seg& rhs) const {
    return t == rhs.t ? n < rhs.n : t < rhs.t;
  }
  void print() const { cout << s << " " << t << " " << n << endl; }
};
bool cmp(const Seg& lhs, const Seg& rhs) { return lhs.s < rhs.s; }
int main() {
  int n, k;
  cin >> n >> k;
  set<Seg> S;
  vector<Seg> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].s >> v[i].t;
    v[i].n = i + 1;
  }
  vector<int> removed;
  sort(v.begin(), v.end(), cmp);
  for (int i = 0; i < (int)v.size(); i++) {
    auto& seg = v[i];
    while (S.size() && S.begin()->t < seg.s) S.erase(S.begin());
    S.insert(seg);
    if ((int)S.size() > k) {
      auto it = S.end();
      --it;
      removed.push_back(it->n);
      S.erase(it);
    }
  }
  sort(removed.begin(), removed.end());
  cout << removed.size() << endl;
  for (auto& it : removed) cout << it << ' ';
  cout << endl;
}

### Prompt

Please formulate 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;
using ll = long long;
struct Seg {
  int s, t, n;
  Seg(int s = 0, int t = 0, int n = 0) : s(s), t(t), n(n) {}
  bool operator<(const Seg& rhs) const {
    return t == rhs.t ? n < rhs.n : t < rhs.t;
  }
  void print() const { cout << s << " " << t << " " << n << endl; }
};
bool cmp(const Seg& lhs, const Seg& rhs) { return lhs.s < rhs.s; }
int main() {
  int n, k;
  cin >> n >> k;
  set<Seg> S;
  vector<Seg> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i].s >> v[i].t;
    v[i].n = i + 1;
  }
  vector<int> removed;
  sort(v.begin(), v.end(), cmp);
  for (int i = 0; i < (int)v.size(); i++) {
    auto& seg = v[i];
    while (S.size() && S.begin()->t < seg.s) S.erase(S.begin());
    S.insert(seg);
    if ((int)S.size() > k) {
      auto it = S.end();
      --it;
      removed.push_back(it->n);
      S.erase(it);
    }
  }
  sort(removed.begin(), removed.end());
  cout << removed.size() << endl;
  for (auto& it : removed) cout << it << ' ';
  cout << endl;
}

```