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 
#include <bits/stdc++.h>
using namespace std;
struct event {
  long long x, d, ind, r;
};
bool cmp(event a, event b) {
  if (a.x != b.x)
    return a.x < b.x;
  else
    return a.d > b.d;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<event> events;
  for (long long i = 0; i < n; i++) {
    long long l, r;
    cin >> l >> r;
    events.push_back({l, 1, i, r});
    events.push_back({r, -1, i, r});
  }
  sort((events).begin(), (events).end(), cmp);
  long long bal = 0;
  set<pair<long long, long long>> order;
  vector<bool> used(n);
  long long ans = 0;
  for (long long i = 0; i < 2 * n; i++) {
    if (!used[events[i].ind]) bal += events[i].d;
    if (events[i].d == -1) order.erase({events[i].r, events[i].ind});
    order.insert({events[i].r, events[i].ind});
    if (bal > k && (i == 2 * n - 1 || events[i + 1].x > events[i].x ||
                    events[i + 1].d != 1)) {
      ans += bal - k;
      long long ctr = 0;
      vector<pair<long long, long long>> ers;
      for (auto j = order.rbegin(); j != order.rend() && ctr < bal - k;
           j++, ctr++) {
        used[j->second] = 1;
        ers.push_back(*j);
      }
      for (auto &j : ers) order.erase(j);
      bal = k;
    }
  }
  cout << ans << '\n';
  for (long long i = 0; i < n; i++)
    if (used[i]) cout << i + 1 << " ";
  return 0;
}
