#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
const int INF = 1e9 + 1;
const int MOD = 1234567;
const long long LINF = 1e18 + 1;
vector<pair<int, int> > p[200001];
set<pair<int, int> > s;
int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    p[l].emplace_back(r, i);
  }
  vector<int> ans;
  for (int i = 1; i <= 200000; i++) {
    for (auto it : p[i]) s.insert(it);
    while (s.size() && s.begin()->first < i) s.erase(s.begin());
    while (s.size() > K) {
      ans.push_back(s.rbegin()->second);
      s.erase(--s.end());
    }
  }
  printf("%d\n", ans.size());
  for (auto a : ans) printf("%d ", a);
  puts("");
  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;
#pragma warning(disable : 4996)
const int INF = 1e9 + 1;
const int MOD = 1234567;
const long long LINF = 1e18 + 1;
vector<pair<int, int> > p[200001];
set<pair<int, int> > s;
int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) {
    int l, r;
    scanf("%d %d", &l, &r);
    p[l].emplace_back(r, i);
  }
  vector<int> ans;
  for (int i = 1; i <= 200000; i++) {
    for (auto it : p[i]) s.insert(it);
    while (s.size() && s.begin()->first < i) s.erase(s.begin());
    while (s.size() > K) {
      ans.push_back(s.rbegin()->second);
      s.erase(--s.end());
    }
  }
  printf("%d\n", ans.size());
  for (auto a : ans) printf("%d ", a);
  puts("");
  return 0;
}

```