#include <bits/stdc++.h>
using namespace std;
vector<pair<pair<int, int>, int> > v;
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.first.second > y.first.second;
}
vector<int> a[201];
vector<int> ele;
int flag[201];
int main() {
  int n, k;
  cin >> n >> k;
  int l = 200, r = 0;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    v.push_back(make_pair(make_pair(x, y), i));
    l = min(l, x);
    r = max(r, y);
  }
  sort(v.begin(), v.end(), cmp);
  for (int i = l; i <= r; i++) {
    for (int j = 0; j < v.size(); j++) {
      if (v[j].first.first <= i && v[j].first.second >= i) {
        a[i].push_back(v[j].second);
      }
    }
    if (a[i].size() > k) {
      ele.push_back(i);
    }
  }
  vector<int> ans;
  for (auto i : ele) {
    int cnt = a[i].size() - k;
    for (int j = 0; j < a[i].size(); j++) {
      if (flag[a[i][j]] == 1) {
        cnt--;
      }
    }
    if (cnt > 0) {
      for (int j = 0; j < a[i].size(); j++) {
        int temp = a[i][j];
        if (flag[temp] == 0 && cnt > 0) {
          flag[temp] = 1;
          ans.push_back(temp);
          cnt--;
        }
      }
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
}

### 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<pair<pair<int, int>, int> > v;
bool cmp(pair<pair<int, int>, int> x, pair<pair<int, int>, int> y) {
  return x.first.second > y.first.second;
}
vector<int> a[201];
vector<int> ele;
int flag[201];
int main() {
  int n, k;
  cin >> n >> k;
  int l = 200, r = 0;
  for (int i = 1; i <= n; i++) {
    int x, y;
    cin >> x >> y;
    v.push_back(make_pair(make_pair(x, y), i));
    l = min(l, x);
    r = max(r, y);
  }
  sort(v.begin(), v.end(), cmp);
  for (int i = l; i <= r; i++) {
    for (int j = 0; j < v.size(); j++) {
      if (v[j].first.first <= i && v[j].first.second >= i) {
        a[i].push_back(v[j].second);
      }
    }
    if (a[i].size() > k) {
      ele.push_back(i);
    }
  }
  vector<int> ans;
  for (auto i : ele) {
    int cnt = a[i].size() - k;
    for (int j = 0; j < a[i].size(); j++) {
      if (flag[a[i][j]] == 1) {
        cnt--;
      }
    }
    if (cnt > 0) {
      for (int j = 0; j < a[i].size(); j++) {
        int temp = a[i][j];
        if (flag[temp] == 0 && cnt > 0) {
          flag[temp] = 1;
          ans.push_back(temp);
          cnt--;
        }
      }
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
}

```