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;
long long int mod1 = 998244353;
long long int mod = 1e9 + 7;
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
vector<pair<int, int>> l[200005];
vector<pair<int, int>> r[200005];
void solve() {
  long long int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int x, y;
    cin >> x >> y;
    l[x].push_back(make_pair(-y, i));
    r[y + 1].push_back(make_pair(-y, i));
  }
  set<pair<int, int>> s;
  int c = 0;
  vector<int> ans;
  for (int i = 1; i <= 200000; ++i) {
    for (int j = 0; j < r[i].size(); ++j)
      if (s.find(r[i][j]) != s.end()) s.erase(r[i][j]);
    for (int j = 0; j < l[i].size(); ++j) s.insert(l[i][j]);
    while (s.size() > k) {
      auto it = s.begin();
      ++c;
      ans.push_back((*it).second);
      s.erase(it);
    }
  }
  cout << c << "\n";
  for (int i = 0; i < c; ++i) cout << ans[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}