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;
void fastInOut();
const int N = 200001;
int n, k, l, r;
vector<pair<int, int> > in[N];
vector<int> ot[N], ret;
set<pair<int, int> > cur;
int main() {
  fastInOut();
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> l >> r;
    in[l].push_back({r, i});
    ot[r].push_back(i);
  }
  for (int i = 1; i < N; ++i) {
    for (auto it : in[i]) cur.insert(it);
    while (int(cur.size()) > k) {
      pair<int, int> lst = *cur.rbegin();
      ret.push_back(lst.second);
      cur.erase(lst);
    }
    for (auto it : ot[i]) cur.erase({i, it});
  }
  sort(ret.begin(), ret.end());
  cout << int(ret.size()) << "\n";
  for (int i = 0; i < int(ret.size()); ++i)
    cout << ret[i] + 1 << " \n"[i == int(ret.size()) - 1];
  return 0;
}
void fastInOut() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
}