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 N = 2e5 + 3;
const int NMAX = 2e5;
int n, k, l, r;
set<pair<int, int>> st;
int le[N], ri[N];
vector<int> vle[N], vri[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  l = 1e9, r = -1e9;
  for (int i = 1; i <= n; i++) {
    cin >> le[i] >> ri[i];
    l = min(l, le[i]), r = max(r, ri[i]);
    vle[le[i]].push_back(i);
    vri[ri[i] + 1].push_back(i);
  }
  vector<int> ans;
  for (int i = l; i <= r; i++) {
    for (auto e : vle[i]) st.insert({ri[e], e});
    for (auto e : vri[i]) st.erase({ri[e], e});
    while ((int)st.size() > k) {
      auto it = st.end();
      it--;
      ans.push_back((*it).second);
      st.erase(it);
    }
  }
  cout << ans.size() << '\n';
  for (auto e : ans) cout << e << ' ';
  cout << '\n';
  return 0;
}