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 int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
int l[maxn], r[maxn];
int n, k;
vector<int> node[maxn];
set<pair<int, int> > s;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    node[l[i]].push_back(i);
    node[r[i] + 1].push_back(-i);
  }
  vector<int> ans;
  for (int i = 1; i <= 2e5; i++) {
    for (int j = 0; j < node[i].size(); j++) {
      int u = node[i][j];
      if (u > 0) {
        s.insert(make_pair(r[u], u));
      } else
        s.erase(make_pair(r[-u], -u));
    }
    while (s.size() > k) {
      set<pair<int, int> >::iterator it = s.end();
      --it;
      ans.push_back(it->second);
      s.erase(*it);
    }
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << abs(ans[i]) << " ";
  }
  cout << endl;
  return 0;
}