#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  vector<set<int> > ox(201);
  vector<int> l(n + 1);
  vector<int> r(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    for (int j = l[i]; j <= r[i]; j++) ox[j].insert(i);
  }
  int maxmulti = k + 1;
  set<int> deleted;
  while (maxmulti > k) {
    maxmulti = 0;
    vector<int> a(n + 1, 0);
    for (int i = 1; i <= 200; i++) {
      if (maxmulti < ox[i].size()) {
        maxmulti = ox[i].size();
        for (int j = 1; j <= n; j++) {
          a[j] = 0;
        }
        for (set<int>::iterator it = ox[i].begin(); it != ox[i].end(); it++) {
          a[*it]++;
        }
      } else if (maxmulti == ox[i].size()) {
        for (set<int>::iterator it = ox[i].begin(); it != ox[i].end(); it++) {
          a[*it]++;
        }
      }
    }
    if (maxmulti <= k) break;
    int maxhot = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] > a[maxhot])
        maxhot = i;
      else if (a[i] == a[maxhot])
        if (r[i] - l[i] > r[maxhot] - l[maxhot]) maxhot = i;
    }
    deleted.insert(maxhot);
    for (int i = l[maxhot]; i <= r[maxhot]; i++) {
      ox[i].erase(ox[i].find(maxhot));
    }
  }
  cout << deleted.size() << endl;
  for (set<int>::iterator it = deleted.begin(); it != deleted.end(); it++) {
    cout << *it << ' ';
  }
  return 0;
}

### 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<set<int> > ox(201);
  vector<int> l(n + 1);
  vector<int> r(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    for (int j = l[i]; j <= r[i]; j++) ox[j].insert(i);
  }
  int maxmulti = k + 1;
  set<int> deleted;
  while (maxmulti > k) {
    maxmulti = 0;
    vector<int> a(n + 1, 0);
    for (int i = 1; i <= 200; i++) {
      if (maxmulti < ox[i].size()) {
        maxmulti = ox[i].size();
        for (int j = 1; j <= n; j++) {
          a[j] = 0;
        }
        for (set<int>::iterator it = ox[i].begin(); it != ox[i].end(); it++) {
          a[*it]++;
        }
      } else if (maxmulti == ox[i].size()) {
        for (set<int>::iterator it = ox[i].begin(); it != ox[i].end(); it++) {
          a[*it]++;
        }
      }
    }
    if (maxmulti <= k) break;
    int maxhot = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] > a[maxhot])
        maxhot = i;
      else if (a[i] == a[maxhot])
        if (r[i] - l[i] > r[maxhot] - l[maxhot]) maxhot = i;
    }
    deleted.insert(maxhot);
    for (int i = l[maxhot]; i <= r[maxhot]; i++) {
      ox[i].erase(ox[i].find(maxhot));
    }
  }
  cout << deleted.size() << endl;
  for (set<int>::iterator it = deleted.begin(); it != deleted.end(); it++) {
    cout << *it << ' ';
  }
  return 0;
}

```