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;
int64_t mod = 1e9 + 7;
int64_t mod1 = 1e9 + 6;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((((a % mod) * (a % mod)) % mod), b / 2) % mod;
  else
    return ((a % mod) * (power((((a % mod) * (a % mod)) % mod), b / 2) % mod)) %
           mod;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t n, k;
  cin >> n >> k;
  vector<pair<int64_t, int64_t>> v;
  map<pair<int64_t, int64_t>, int64_t> m;
  map<pair<int64_t, int64_t>, int64_t> m1;
  for (int64_t i = 1; i <= n; ++i) {
    int64_t x, y;
    cin >> x >> y;
    m[{y, i}] = x;
    m1[{x, i}] = y;
    v.push_back({x, -i});
    v.push_back({y, i});
  }
  sort(v.begin(), v.end());
  set<pair<int64_t, int64_t>> s;
  vector<int64_t> ans;
  bool vis[n + 1];
  memset(vis, false, sizeof(vis));
  for (int64_t i = 0; i < v.size(); ++i) {
    if (v[i].second > 0) {
      if (vis[v[i].second]) continue;
      int64_t in = v[i].second;
      int64_t pt = v[i].first;
      if (int64_t(s.size()) > k) {
        int64_t extra = int64_t(s.size()) - k;
        auto it = s.lower_bound({pt, in});
        s.erase(it);
        while (extra) {
          auto it = s.end();
          it--;
          pair<int64_t, int64_t> p = *it;
          ans.push_back(p.second);
          vis[p.second] = true;
          s.erase(it);
          extra--;
        }
      } else {
        auto it = s.lower_bound({pt, in});
        s.erase(it);
      }
    } else {
      s.insert({m1[{v[i].first, abs(v[i].second)}], abs(v[i].second)});
    }
  }
  sort(ans.begin(), ans.end());
  cout << int64_t(ans.size()) << '\n';
  for (auto it : ans) cout << it << " ";
}