#include <bits/stdc++.h>
using namespace std;
int N, K, numRemoved[1000 + 3], numCover[1000 + 3];
pair<int, int> segments[1000 + 3];
vector<int> events[1000 + 3];
vector<int> removedSegments;
set<pair<int, int> > activeSegments;
int main() {
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    int l, r;
    cin >> l >> r;
    segments[i] = {l, r};
    events[l].push_back(i);
    events[r + 1].push_back(-i);
    numCover[l] += 1;
    numCover[r + 1] -= 1;
  }
  memset(numRemoved, 0, sizeof(numRemoved));
  for (int i = 1; i <= 1000; i++) {
    numCover[i] += numCover[i - 1];
    numRemoved[i] += numRemoved[i - 1];
    for (auto idx : events[i]) {
      if (idx > 0) {
        activeSegments.insert({segments[idx].second, idx});
      } else {
        set<pair<int, int> >::iterator ptr =
            activeSegments.find({segments[-idx].second, -idx});
        if (ptr != activeSegments.end()) activeSegments.erase(ptr);
      }
    }
    int countToBeRemoved = (numCover[i] - numRemoved[i]) - K;
    while (countToBeRemoved > 0) {
      pair<int, int> toBeRemoved = *(--activeSegments.end());
      removedSegments.push_back(toBeRemoved.second);
      numRemoved[i] += 1;
      numRemoved[toBeRemoved.first + 1] -= 1;
      activeSegments.erase(activeSegments.find(toBeRemoved));
      countToBeRemoved -= 1;
    }
  }
  cout << removedSegments.size() << endl;
  for (auto id : removedSegments) cout << id << " ";
  cout << endl;
}

### Prompt

Create a solution in CPP for 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;
int N, K, numRemoved[1000 + 3], numCover[1000 + 3];
pair<int, int> segments[1000 + 3];
vector<int> events[1000 + 3];
vector<int> removedSegments;
set<pair<int, int> > activeSegments;
int main() {
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    int l, r;
    cin >> l >> r;
    segments[i] = {l, r};
    events[l].push_back(i);
    events[r + 1].push_back(-i);
    numCover[l] += 1;
    numCover[r + 1] -= 1;
  }
  memset(numRemoved, 0, sizeof(numRemoved));
  for (int i = 1; i <= 1000; i++) {
    numCover[i] += numCover[i - 1];
    numRemoved[i] += numRemoved[i - 1];
    for (auto idx : events[i]) {
      if (idx > 0) {
        activeSegments.insert({segments[idx].second, idx});
      } else {
        set<pair<int, int> >::iterator ptr =
            activeSegments.find({segments[-idx].second, -idx});
        if (ptr != activeSegments.end()) activeSegments.erase(ptr);
      }
    }
    int countToBeRemoved = (numCover[i] - numRemoved[i]) - K;
    while (countToBeRemoved > 0) {
      pair<int, int> toBeRemoved = *(--activeSegments.end());
      removedSegments.push_back(toBeRemoved.second);
      numRemoved[i] += 1;
      numRemoved[toBeRemoved.first + 1] -= 1;
      activeSegments.erase(activeSegments.find(toBeRemoved));
      countToBeRemoved -= 1;
    }
  }
  cout << removedSegments.size() << endl;
  for (auto id : removedSegments) cout << id << " ";
  cout << endl;
}

```