#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
const int N = 2e5 + 1;
const int oo = 1e9;
const double pi = acos(-1);
struct seg {
  int left, right, idx;
};
bool comp1(seg a, seg b) {
  return a.left < b.left || (a.left == b.left && a.right < b.right) ||
         (a.left == b.left && a.right == b.right && a.idx < b.idx);
}
bool comp2(seg a, seg b) {
  return a.right < b.right || (a.right == b.right && a.left < b.left) ||
         (a.right == b.right && a.left == b.left && a.idx < b.idx);
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<seg> minL(n), minR(n);
  for (int i = 0; i < n; ++i) {
    cin >> minL[i].left >> minL[i].right;
    minL[i].idx = i + 1;
    minR[i] = minL[i];
  }
  sort(minL.begin(), minL.end(), comp1);
  sort(minR.begin(), minR.end(), comp2);
  vector<int> ans;
  set<pair<pair<int, int>, int>> active;
  for (int p = 1, i = 0, j = 0; p < N; ++p) {
    while (i < n && minL[i].left <= p) {
      active.insert({{minL[i].right, minL[i].left}, minL[i].idx});
      ++i;
    }
    while (active.size() > k) {
      auto best = active.end();
      --best;
      ans.push_back(best->second);
      active.erase(best);
    }
    while (j < n && minR[j].right <= p) {
      active.erase({{minR[j].right, minR[j].left}, minR[j].idx});
      ++j;
    }
  }
  cout << ans.size() << endl;
  for (auto a : ans) cout << a << " ";
  cout << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const int M = 1e9 + 7;
const int N = 2e5 + 1;
const int oo = 1e9;
const double pi = acos(-1);
struct seg {
  int left, right, idx;
};
bool comp1(seg a, seg b) {
  return a.left < b.left || (a.left == b.left && a.right < b.right) ||
         (a.left == b.left && a.right == b.right && a.idx < b.idx);
}
bool comp2(seg a, seg b) {
  return a.right < b.right || (a.right == b.right && a.left < b.left) ||
         (a.right == b.right && a.left == b.left && a.idx < b.idx);
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<seg> minL(n), minR(n);
  for (int i = 0; i < n; ++i) {
    cin >> minL[i].left >> minL[i].right;
    minL[i].idx = i + 1;
    minR[i] = minL[i];
  }
  sort(minL.begin(), minL.end(), comp1);
  sort(minR.begin(), minR.end(), comp2);
  vector<int> ans;
  set<pair<pair<int, int>, int>> active;
  for (int p = 1, i = 0, j = 0; p < N; ++p) {
    while (i < n && minL[i].left <= p) {
      active.insert({{minL[i].right, minL[i].left}, minL[i].idx});
      ++i;
    }
    while (active.size() > k) {
      auto best = active.end();
      --best;
      ans.push_back(best->second);
      active.erase(best);
    }
    while (j < n && minR[j].right <= p) {
      active.erase({{minR[j].right, minR[j].left}, minR[j].idx});
      ++j;
    }
  }
  cout << ans.size() << endl;
  for (auto a : ans) cout << a << " ";
  cout << endl;
  return 0;
}

```