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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int i, n, k, x, y;
  cin >> n >> k;
  vector<pair<pair<int, int>, int>> a;
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    a.push_back({{x, y}, i + 1});
  }
  sort(a.begin(), a.end());
  vector<int> ans;
  set<pair<int, int>> second;
  int ind = 0;
  for (i = 1; i < 200005; i++) {
    while (ind < n and a[ind].first.first <= i) {
      second.insert({a[ind].first.second, a[ind].second});
      ind++;
    }
    while (!second.empty() and (second.begin()->first < i)) {
      second.erase(second.begin());
    }
    while (second.size() > k) {
      ans.push_back(prev(second.end())->second);
      second.erase(prev(second.end()));
    }
  }
  cout << ans.size();
  cout << "\n";
  for (auto i : ans) cout << i << " ";
  return 0;
}