#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast")
constexpr ll INF = 1e18;
constexpr int inf = 1e9;
constexpr ll mod = 1000000007;
constexpr ll mod2 = 998244353;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
using tup = tuple<int, int, int>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, K;
  cin >> N >> K;
  vector<vector<P>> pos(200005);
  for (int i = 0; i < N; i++) {
    int l, r;
    cin >> l >> r;
    pos[l].emplace_back(r, i);
  }
  vector<int> ans;
  vector<int> r(200005, 0);
  int now = 0;
  priority_queue<P> que;
  for (int i = 1; i <= 200000; i++) {
    now += pos[i].size();
    for (int j = 0; j < pos[i].size(); j++) {
      que.emplace(pos[i][j]);
      r[pos[i][j].first]++;
    }
    while (now > K) {
      ans.emplace_back(que.top().second);
      r[que.top().first]--;
      que.pop();
      now--;
    }
    now -= r[i];
  }
  cout << ans.size() << "\n";
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << " \n"[i + 1 == ans.size()];
  }
  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;
using ll = long long;
using P = pair<int, int>;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast")
constexpr ll INF = 1e18;
constexpr int inf = 1e9;
constexpr ll mod = 1000000007;
constexpr ll mod2 = 998244353;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
using tup = tuple<int, int, int>;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, K;
  cin >> N >> K;
  vector<vector<P>> pos(200005);
  for (int i = 0; i < N; i++) {
    int l, r;
    cin >> l >> r;
    pos[l].emplace_back(r, i);
  }
  vector<int> ans;
  vector<int> r(200005, 0);
  int now = 0;
  priority_queue<P> que;
  for (int i = 1; i <= 200000; i++) {
    now += pos[i].size();
    for (int j = 0; j < pos[i].size(); j++) {
      que.emplace(pos[i][j]);
      r[pos[i][j].first]++;
    }
    while (now > K) {
      ans.emplace_back(que.top().second);
      r[que.top().first]--;
      que.pop();
      now--;
    }
    now -= r[i];
  }
  cout << ans.size() << "\n";
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << " \n"[i + 1 == ans.size()];
  }
  return 0;
}

```