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;
vector<pair<long long int, long long int> > in[200005], out[200005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k, i, j, l, r;
  cin >> n >> k;
  set<pair<long long int, long long int> > st;
  vector<long long int> ans;
  for (i = 0; i < n; i++) {
    cin >> l >> r;
    in[l].push_back({r, i});
    out[r + 1].push_back({r, i});
  }
  for (i = 1; i <= 200004; i++) {
    for (j = 0; j < in[i].size(); j++) {
      st.insert(in[i][j]);
    }
    for (j = 0; j < out[i].size(); j++) {
      auto it = st.lower_bound(out[i][j]);
      if (it == st.end()) {
        continue;
      }
      if (it->first == out[i][j].first && it->second == out[i][j].second) {
        st.erase(it);
      }
    }
    if (st.size() > k) {
      while (st.size() > k) {
        auto it = st.end();
        it--;
        ans.push_back(it->second);
        st.erase(it);
      }
    }
  }
  cout << ans.size() << endl;
  for (i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << " ";
  }
}