#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> a(n);
  int x, y;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    a[i].first = y;
    a[i].second = x;
  }
  vector<pair<int, int>> b = a;
  sort(a.begin(), a.end());
  vector<pair<int, int>> ans;
  int m = n;
  for (int i = 1; i < 200; i++) {
    int t = 0;
    for (int j = 0; j < n; j++) {
      if (i >= a[j].second && i <= a[j].first) {
        t++;
      }
    }
    for (int j = m - 1; j > -1; j--) {
      if (t <= k) break;
      if (a[j].first < i) break;
      if (a[j].second <= i) {
        ans.push_back(a[j]);
        a[j].second = 200 + 1;
        t--;
      }
    }
  }
  vector<bool> used(200 + 1, false);
  vector<int> ans_ind;
  for (auto j : ans) {
    for (int i = 0; i < n; i++) {
      if (!used[i] && b[i] == j) {
        used[i] = true;
        ans_ind.push_back(i + 1);
        break;
      }
    }
  }
  cout << ans_ind.size() << '\n';
  for (auto i : ans_ind) {
    cout << i << ' ';
  }
}

### Prompt

Please create a solution in CPP 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;
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int>> a(n);
  int x, y;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    a[i].first = y;
    a[i].second = x;
  }
  vector<pair<int, int>> b = a;
  sort(a.begin(), a.end());
  vector<pair<int, int>> ans;
  int m = n;
  for (int i = 1; i < 200; i++) {
    int t = 0;
    for (int j = 0; j < n; j++) {
      if (i >= a[j].second && i <= a[j].first) {
        t++;
      }
    }
    for (int j = m - 1; j > -1; j--) {
      if (t <= k) break;
      if (a[j].first < i) break;
      if (a[j].second <= i) {
        ans.push_back(a[j]);
        a[j].second = 200 + 1;
        t--;
      }
    }
  }
  vector<bool> used(200 + 1, false);
  vector<int> ans_ind;
  for (auto j : ans) {
    for (int i = 0; i < n; i++) {
      if (!used[i] && b[i] == j) {
        used[i] = true;
        ans_ind.push_back(i + 1);
        break;
      }
    }
  }
  cout << ans_ind.size() << '\n';
  for (auto i : ans_ind) {
    cout << i << ' ';
  }
}

```