#include <bits/stdc++.h>
using namespace std;
vector<int> cover[201];
vector<int> ans;
int main() {
  int N, K;
  cin >> N >> K;
  for (int seg = 1; N; --N, ++seg) {
    int l, r;
    cin >> l >> r;
    for (int idx = l; idx <= r; ++idx) cover[idx].push_back(seg);
  }
  for (int point = 1; point <= 200; ++point)
    if (cover[point].size() > K) {
      vector<int> freq(201);
      for (const int &itm : cover[point]) ++freq[itm];
      for (int idx = 1; idx <= 200; ++idx)
        if (idx != point && cover[idx].size() > K)
          for (const int &itm : cover[idx])
            if (freq[itm]) ++freq[itm];
      int seg = 0;
      for (int idx = 1; idx <= 200; ++idx)
        if (freq[idx] > freq[seg]) seg = idx;
      for (int idx = 1; idx <= 200; ++idx) {
        bool found = false;
        vector<int>::iterator itFound;
        for (auto it = cover[idx].begin(); it != cover[idx].end(); ++it)
          if (seg == *it) {
            found = true;
            itFound = it;
            break;
          }
        if (found) cover[idx].erase(itFound);
      }
      ans.push_back(seg);
      if (cover[point].size() > K) --point;
    }
  cout << ans.size() << '\n';
  for (const int &itm : ans) cout << itm << ' ';
}

### 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;
vector<int> cover[201];
vector<int> ans;
int main() {
  int N, K;
  cin >> N >> K;
  for (int seg = 1; N; --N, ++seg) {
    int l, r;
    cin >> l >> r;
    for (int idx = l; idx <= r; ++idx) cover[idx].push_back(seg);
  }
  for (int point = 1; point <= 200; ++point)
    if (cover[point].size() > K) {
      vector<int> freq(201);
      for (const int &itm : cover[point]) ++freq[itm];
      for (int idx = 1; idx <= 200; ++idx)
        if (idx != point && cover[idx].size() > K)
          for (const int &itm : cover[idx])
            if (freq[itm]) ++freq[itm];
      int seg = 0;
      for (int idx = 1; idx <= 200; ++idx)
        if (freq[idx] > freq[seg]) seg = idx;
      for (int idx = 1; idx <= 200; ++idx) {
        bool found = false;
        vector<int>::iterator itFound;
        for (auto it = cover[idx].begin(); it != cover[idx].end(); ++it)
          if (seg == *it) {
            found = true;
            itFound = it;
            break;
          }
        if (found) cover[idx].erase(itFound);
      }
      ans.push_back(seg);
      if (cover[point].size() > K) --point;
    }
  cout << ans.size() << '\n';
  for (const int &itm : ans) cout << itm << ' ';
}

```