#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MAXN = 200005;
vector<int> add[MAXN], rem[MAXN], ret;
set<pii> G;
int n, k, r[MAXN], l[MAXN];
bool visted[MAXN];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  while (cin >> n >> k) {
    ret.clear();
    G.clear();
    memset(visted, 0, sizeof(visted));
    for (int i = 1; i <= n; ++i) {
      cin >> l[i] >> r[i];
      add[l[i]].push_back(i);
      rem[r[i] + 1].push_back(i);
    }
    for (int i = 1; i <= 200000; ++i) {
      for (auto x : rem[i]) {
        if (!visted[x]) {
          G.erase(G.find(make_pair(r[x], x)));
        }
      }
      for (auto x : add[i]) {
        G.insert(make_pair(r[x], x));
      }
      while (G.size() > k) {
        auto p = *prev(G.end());
        G.erase(G.find(p));
        visted[p.second] = true;
        ret.push_back(p.second);
      }
    }
    cout << ret.size() << '\n';
    for (auto x : ret) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
using pii = pair<int, int>;
const int MAXN = 200005;
vector<int> add[MAXN], rem[MAXN], ret;
set<pii> G;
int n, k, r[MAXN], l[MAXN];
bool visted[MAXN];
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  while (cin >> n >> k) {
    ret.clear();
    G.clear();
    memset(visted, 0, sizeof(visted));
    for (int i = 1; i <= n; ++i) {
      cin >> l[i] >> r[i];
      add[l[i]].push_back(i);
      rem[r[i] + 1].push_back(i);
    }
    for (int i = 1; i <= 200000; ++i) {
      for (auto x : rem[i]) {
        if (!visted[x]) {
          G.erase(G.find(make_pair(r[x], x)));
        }
      }
      for (auto x : add[i]) {
        G.insert(make_pair(r[x], x));
      }
      while (G.size() > k) {
        auto p = *prev(G.end());
        G.erase(G.find(p));
        visted[p.second] = true;
        ret.push_back(p.second);
      }
    }
    cout << ret.size() << '\n';
    for (auto x : ret) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}

```