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;
const long long INF = (long long)1 << 62;
const long long MOD = 1e9 + 7;
const int iINF = 1 << 30;
const double PI = 3.14159265359;
const int MAXP = 200005;
int n, k;
vector<pair<int, int> > intv;
vector<int> ent[MAXP];
vector<int> ext[MAXP];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  intv.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> intv[i].first >> intv[i].second;
    ent[intv[i].first].push_back(i);
    ext[intv[i].second + 1].push_back(i);
  }
  set<pair<int, int> > curr;
  vector<int> ans;
  for (int p = 1; p < MAXP; p++) {
    for (int i = 0; i < ext[p].size(); i++) {
      curr.erase({intv[ext[p][i]].second, ext[p][i]});
    }
    for (int i = 0; i < ent[p].size(); i++) {
      curr.insert({intv[ent[p][i]].second, ent[p][i]});
    }
    while (curr.size() > k) {
      auto it = curr.end();
      it--;
      ans.push_back((*it).second);
      curr.erase(it);
    }
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
  cout << "\n";
  return 0;
}