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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long block;
int mp[1000001];
long long powe(long long n) {
  if (n == 0) return 1;
  long long p = powe(n / 2);
  p = p * p;
  if (n % 2 == 1) p *= 3;
  return p;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  vector<pair<int, pair<int, int>>> pts(n);
  for (int i = 0; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    pts[i] = {x, {y, i}};
  }
  sort(pts.begin(), pts.end(),
       [](pair<int, pair<int, int>> a, pair<int, pair<int, int>> b) {
         if (a.first != b.first) return a.first < b.first;
         return a.second.first < b.second.first;
       });
  bool vis[n];
  memset(vis, 0, sizeof(vis));
  int c = 0, it = 0;
  vector<int> ans;
  priority_queue<pair<int, int>> pq;
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      mpq;
  for (int i = 0; i < 200005; ++i) {
    while (it < n && pts[it].first == i) {
      c++;
      pq.push({pts[it].second.first, pts[it].second.second});
      mpq.push({pts[it].second.first, pts[it].second.second});
      ++it;
    }
    while (mpq.size() && mpq.top().first < i) {
      int in = mpq.top().second;
      if (vis[in] == false) {
        c--;
        vis[in] = true;
      }
      mpq.pop();
    }
    while (c > k && pq.size()) {
      int in = pq.top().second;
      if (vis[in] == false) {
        c--;
        vis[in] = true;
        ans.push_back(in + 1);
      }
      pq.pop();
    }
  }
  cout << ans.size() << endl;
  for (auto l : ans) cout << l << " ";
  cout << endl;
}