#include <bits/stdc++.h>
using namespace std;
const int MOD = int(1e9) + 7;
const long long int MOD64 = (long long int)(1e18) + 7;
const int INF = 0x7fffffff;
const long long int INF64 = 0x7fffffffffffffff;
const int N = 2e5 + 10;
vector<pair<pair<int, int>, int> > p;
vector<int> v[N];
vector<int> num;
vector<int> ans;
bool vis[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, n;
  cin >> n >> m;
  int a, b, c;
  int r = 0;
  for (int i = int(0); i < int(n); i++) {
    cin >> a >> b;
    p.push_back(make_pair(make_pair(b, a), i + 1));
    r = max(r, b);
  }
  sort((p).begin(), (p).end());
  memset(vis, false, sizeof(vis));
  for (int i = int(0); i < int(n); i++) {
    a = p[i].second;
    b = p[i].first.first;
    c = p[i].first.second;
    for (int j = int(c); j < int(b + 1); j++) {
      v[j].push_back(a);
    }
  }
  for (int i = int(1); i < int(r + 1); i++) {
    if ((int)(v[i]).size() > m) {
      int sum = (int)(v[i]).size() - m;
      for (int j = int(0); j < int((int)(v[i]).size()); j++) {
        if (vis[v[i][j]]) sum--;
      }
      for (int j = (int)(v[i]).size() - 1; j >= 0; j--) {
        if (sum <= 0) break;
        if (vis[v[i][j]] == false) {
          ans.push_back(v[i][j]);
          vis[v[i][j]] = true;
          sum--;
        }
      }
    }
  }
  cout << (int)(ans).size() << endl;
  sort((ans).begin(), (ans).end());
  for (int i = int(0); i < int((int)(ans).size()); i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

### 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;
const int MOD = int(1e9) + 7;
const long long int MOD64 = (long long int)(1e18) + 7;
const int INF = 0x7fffffff;
const long long int INF64 = 0x7fffffffffffffff;
const int N = 2e5 + 10;
vector<pair<pair<int, int>, int> > p;
vector<int> v[N];
vector<int> num;
vector<int> ans;
bool vis[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, n;
  cin >> n >> m;
  int a, b, c;
  int r = 0;
  for (int i = int(0); i < int(n); i++) {
    cin >> a >> b;
    p.push_back(make_pair(make_pair(b, a), i + 1));
    r = max(r, b);
  }
  sort((p).begin(), (p).end());
  memset(vis, false, sizeof(vis));
  for (int i = int(0); i < int(n); i++) {
    a = p[i].second;
    b = p[i].first.first;
    c = p[i].first.second;
    for (int j = int(c); j < int(b + 1); j++) {
      v[j].push_back(a);
    }
  }
  for (int i = int(1); i < int(r + 1); i++) {
    if ((int)(v[i]).size() > m) {
      int sum = (int)(v[i]).size() - m;
      for (int j = int(0); j < int((int)(v[i]).size()); j++) {
        if (vis[v[i][j]]) sum--;
      }
      for (int j = (int)(v[i]).size() - 1; j >= 0; j--) {
        if (sum <= 0) break;
        if (vis[v[i][j]] == false) {
          ans.push_back(v[i][j]);
          vis[v[i][j]] = true;
          sum--;
        }
      }
    }
  }
  cout << (int)(ans).size() << endl;
  sort((ans).begin(), (ans).end());
  for (int i = int(0); i < int((int)(ans).size()); i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}

```