#include <bits/stdc++.h>
using namespace std;
long double pi = 3.14159265358979323;
const double EPS = 1e-12;
const int N = 2e2 + 3;
const int mod = 1e9 + 7;
set<pair<long long, long long>> v[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    for (int j = a; j <= b; j++) {
      v[j].insert({b, i + 1});
    }
  }
  set<long long> ans;
  for (int i = 0; i <= 200; i++) {
    if (v[i].size() <= k) continue;
    auto it = v[i].begin();
    while (it != v[i].end()) {
      auto itr = it;
      itr++;
      if (ans.find(it->second) != ans.end()) {
        v[i].erase(it);
      }
      it = itr;
    }
    if (v[i].size() > k) {
      int cnt = (long long)v[i].size() - k;
      for (auto it = v[i].rbegin(); it != v[i].rend(); it++) {
        ans.insert(it->second);
        cnt--;
        if (!cnt) break;
      }
    }
  }
  cout << (long long)ans.size() << "\n";
  for (auto it : ans) cout << it << " ";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 double pi = 3.14159265358979323;
const double EPS = 1e-12;
const int N = 2e2 + 3;
const int mod = 1e9 + 7;
set<pair<long long, long long>> v[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    for (int j = a; j <= b; j++) {
      v[j].insert({b, i + 1});
    }
  }
  set<long long> ans;
  for (int i = 0; i <= 200; i++) {
    if (v[i].size() <= k) continue;
    auto it = v[i].begin();
    while (it != v[i].end()) {
      auto itr = it;
      itr++;
      if (ans.find(it->second) != ans.end()) {
        v[i].erase(it);
      }
      it = itr;
    }
    if (v[i].size() > k) {
      int cnt = (long long)v[i].size() - k;
      for (auto it = v[i].rbegin(); it != v[i].rend(); it++) {
        ans.insert(it->second);
        cnt--;
        if (!cnt) break;
      }
    }
  }
  cout << (long long)ans.size() << "\n";
  for (auto it : ans) cout << it << " ";
  return 0;
}

```