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 cnt[200005];
vector<pair<int, int> > v[200005];
int main() {
  int n, k, l, r;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> l >> r;
    cnt[l]++, cnt[r + 1]--;
    v[l].push_back({r, i}), v[r + 1].push_back({r, -i});
  }
  set<pair<int, int>, greater<pair<int, int> > > st;
  set<int> ans;
  for (int i = 1; i <= 200000; i++) {
    cnt[i] += cnt[i - 1];
    for (auto x : v[i]) {
      if (x.second > 0)
        st.insert(x);
      else
        st.erase(x);
    }
    while (cnt[i] > k) {
      auto now = st.lower_bound({200005, i});
      cnt[i]--;
      cnt[(*now).first + 1]++;
      ans.insert((*now).second);
      st.erase(now);
    }
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << ' ';
}