#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long a[N], b[N];
vector<vector<long long> > v;
set<long long> op;
set<pair<long long, long long> > en;
vector<long long> ans;
void run() {
  long long n, k, i;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    vector<long long> temp;
    temp.push_back(a[i]);
    temp.push_back(0);
    temp.push_back(i);
    v.push_back(temp);
    temp.clear();
    temp.push_back(b[i]);
    temp.push_back(1);
    temp.push_back(i);
    v.push_back(temp);
  }
  sort(v.begin(), v.end());
  for (auto it : v) {
    if ((long long)op.size() == k + 1) {
      ans.push_back((*en.rbegin()).second);
      en.erase(*en.rbegin());
      op.erase(ans.back());
    }
    if (it[1] == 1) {
      op.erase(it[2]);
      en.erase(make_pair(b[it[2]], it[2]));
    } else {
      op.insert(it[2]);
      en.insert(make_pair(b[it[2]], it[2]));
    }
  }
  cout << (long long)ans.size() << "\n";
  for (auto it : ans) cout << it << " ";
}
void setup() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
}
signed main() {
  setup();
  int tests = 1;
  for (int i = 1; i <= tests; i++) run();
  return 0;
}

### Prompt

Your task is to create 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 N = 2e5 + 5;
long long a[N], b[N];
vector<vector<long long> > v;
set<long long> op;
set<pair<long long, long long> > en;
vector<long long> ans;
void run() {
  long long n, k, i;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
    vector<long long> temp;
    temp.push_back(a[i]);
    temp.push_back(0);
    temp.push_back(i);
    v.push_back(temp);
    temp.clear();
    temp.push_back(b[i]);
    temp.push_back(1);
    temp.push_back(i);
    v.push_back(temp);
  }
  sort(v.begin(), v.end());
  for (auto it : v) {
    if ((long long)op.size() == k + 1) {
      ans.push_back((*en.rbegin()).second);
      en.erase(*en.rbegin());
      op.erase(ans.back());
    }
    if (it[1] == 1) {
      op.erase(it[2]);
      en.erase(make_pair(b[it[2]], it[2]));
    } else {
      op.insert(it[2]);
      en.insert(make_pair(b[it[2]], it[2]));
    }
  }
  cout << (long long)ans.size() << "\n";
  for (auto it : ans) cout << it << " ";
}
void setup() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
}
signed main() {
  setup();
  int tests = 1;
  for (int i = 1; i <= tests; i++) run();
  return 0;
}

```