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>
const int len = 250000;
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<int> p(len, 0), ans;
  vector<pair<int, int> > a(len);
  vector<vector<int> > arr(len);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
    p[a[i].first]++;
    p[a[i].second + 1]--;
    arr[a[i].first].push_back(i);
    arr[a[i].second + 1].push_back(-i);
  }
  for (int i = 1; i < len; i++) {
    p[i] += p[i - 1];
  }
  multiset<int> del;
  int cnt = 0;
  set<pair<int, int>, greater<pair<int, int> > > s;
  for (int i = 1; i < len; i++) {
    if (del.find(i) != del.end()) {
      cnt -= del.count(i);
    }
    for (int j : arr[i]) {
      if (j > 0) {
        s.insert({a[j].second, j});
      } else {
        if (s.find({a[-j].second, -j}) != s.end()) {
          s.erase({a[-j].second, -j});
        }
      }
    }
    if (p[i] - cnt > k) {
      while (p[i] - k - cnt > 0) {
        pair<int, int> t = *s.begin();
        s.erase(s.begin());
        del.insert(t.first + 1);
        ans.push_back(t.second);
        cnt++;
      }
    }
  }
  cout << ans.size() << '\n';
  for (int i : ans) {
    cout << i << ' ';
  }
  return 0;
}