#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000007;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, seg[202] = {0};
  cin >> n >> k;
  pair<long long int, long long int> a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    seg[a[i].first]++;
    seg[a[i].second + 1]--;
  }
  vector<long long int> v;
  for (long long int i = 2; i <= 200; i++) {
    seg[i] += seg[i - 1];
  }
  set<long long int> vis;
  for (long long int i = 1; i <= 200; i++) {
    while (seg[i] > k) {
      long long int idx = -1;
      for (long long int j = 0; j < n; j++) {
        if (vis.count(j)) continue;
        if (idx == -1 && a[j].first <= i)
          idx = j;
        else if (a[j].second - i > a[idx].second - i && a[j].first <= i)
          idx = j;
      }
      for (long long int j = a[idx].first; j <= a[idx].second; j++) {
        seg[j]--;
      }
      v.push_back(idx + 1);
      vis.insert(idx);
    }
  }
  cout << v.size() << endl;
  for (long long int i : v) cout << i << " ";
  cout << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long int mod = 1000000007;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k, seg[202] = {0};
  cin >> n >> k;
  pair<long long int, long long int> a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
    seg[a[i].first]++;
    seg[a[i].second + 1]--;
  }
  vector<long long int> v;
  for (long long int i = 2; i <= 200; i++) {
    seg[i] += seg[i - 1];
  }
  set<long long int> vis;
  for (long long int i = 1; i <= 200; i++) {
    while (seg[i] > k) {
      long long int idx = -1;
      for (long long int j = 0; j < n; j++) {
        if (vis.count(j)) continue;
        if (idx == -1 && a[j].first <= i)
          idx = j;
        else if (a[j].second - i > a[idx].second - i && a[j].first <= i)
          idx = j;
      }
      for (long long int j = a[idx].first; j <= a[idx].second; j++) {
        seg[j]--;
      }
      v.push_back(idx + 1);
      vis.insert(idx);
    }
  }
  cout << v.size() << endl;
  for (long long int i : v) cout << i << " ";
  cout << endl;
  return 0;
}

```