#include <bits/stdc++.h>
using namespace std;
set<pair<int, int> > s;
int k, n;
int sz[200005];
pair<pair<int, int>, int> p[200005];
int ans[200005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, l, r; i <= n; ++i) {
    scanf("%d%d", &l, &r);
    p[i].first = {l, r + 1};
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1, pos = 1; i <= 200000; ++i) {
    sz[i] += sz[i - 1];
    while (!s.empty()) {
      auto it = s.end();
      --it;
      if (it->first != i) break;
      s.erase(it);
    }
    while (pos <= n && p[pos].first.first == i) {
      s.insert({p[pos].first.second, p[pos].second});
      ++sz[i];
      --sz[p[pos].first.second];
      ++pos;
    }
    while (sz[i] > k) {
      auto it = s.end();
      --it;
      --sz[i];
      ans[it->second] = 1;
      ++sz[it->first];
      s.erase(it);
    }
  }
  int tm = 0;
  for (int i = 1; i <= n; ++i) tm += ans[i];
  printf("%d\n", tm);
  for (int i = 1; i <= n; ++i) {
    if (ans[i]) {
      --tm;
      printf("%d%c", i, " \n"[tm == 0]);
    }
  }
  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;
set<pair<int, int> > s;
int k, n;
int sz[200005];
pair<pair<int, int>, int> p[200005];
int ans[200005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, l, r; i <= n; ++i) {
    scanf("%d%d", &l, &r);
    p[i].first = {l, r + 1};
    p[i].second = i;
  }
  sort(p + 1, p + n + 1);
  for (int i = 1, pos = 1; i <= 200000; ++i) {
    sz[i] += sz[i - 1];
    while (!s.empty()) {
      auto it = s.end();
      --it;
      if (it->first != i) break;
      s.erase(it);
    }
    while (pos <= n && p[pos].first.first == i) {
      s.insert({p[pos].first.second, p[pos].second});
      ++sz[i];
      --sz[p[pos].first.second];
      ++pos;
    }
    while (sz[i] > k) {
      auto it = s.end();
      --it;
      --sz[i];
      ans[it->second] = 1;
      ++sz[it->first];
      s.erase(it);
    }
  }
  int tm = 0;
  for (int i = 1; i <= n; ++i) tm += ans[i];
  printf("%d\n", tm);
  for (int i = 1; i <= n; ++i) {
    if (ans[i]) {
      --tm;
      printf("%d%c", i, " \n"[tm == 0]);
    }
  }
  return 0;
}

```