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;
struct SegTree {
  vector<int64_t> changed;
  SegTree(int len) : changed(4 * len, 0) {}
  void Push(int v, int vl, int vr) {
    if (changed[v] == 0) {
      return;
    }
    if (vl != vr) {
      changed[2 * v] += changed[v];
      changed[2 * v + 1] += changed[v];
      changed[v] = 0;
    }
  }
  void SumImpl(int l, int r, int delta, int v, int vl, int vr) {
    Push(v, vl, vr);
    if (l > r) {
      return;
    }
    if (l == vl && r == vr) {
      changed[v] += delta;
      Push(v, l, r);
      return;
    }
    int m = (vl + vr) / 2;
    SumImpl(l, min(r, m), delta, 2 * v, vl, m);
    SumImpl(max(l, m + 1), r, delta, 2 * v + 1, m + 1, vr);
  }
  void Sum(int l, int r, int delta) {
    return SumImpl(l, r, delta, 1, 0, changed.size() / 4 - 1);
  }
  int64_t GetImpl(int p, int v, int vl, int vr) {
    Push(v, vl, vr);
    if (vl == vr) {
      return changed[v];
    }
    int m = (vl + vr) / 2;
    if (m < p) {
      return GetImpl(p, 2 * v + 1, m + 1, vr);
    }
    return GetImpl(p, 2 * v, vl, m);
  }
  int64_t Get(int p) { return GetImpl(p, 1, 0, changed.size() / 4 - 1); }
};
template <class T>
using max_heap = priority_queue<T, vector<T>, less<T>>;
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int count, bound;
  cin >> count >> bound;
  int max_len = 200002;
  SegTree tree(max_len);
  vector<vector<pair<int, int>>> events(max_len);
  vector<pair<int, int>> segs(count);
  for (int i = 0; i < segs.size(); ++i) {
    cin >> segs[i].first >> segs[i].second;
    tree.Sum(segs[i].first, segs[i].second, 1);
    events[segs[i].first].emplace_back(i, 1);
    events[segs[i].second + 1].emplace_back(i, -1);
  }
  set<int> deleted;
  set<int> actives;
  max_heap<pair<int, int>> hp;
  for (int p = 0; p < max_len; ++p) {
    for (auto&& [i, t] : events[p]) {
      if (t == 1) {
        actives.insert(i);
        hp.push({segs[i].second, i});
      } else if (t == -1) {
        actives.erase(i);
      }
    }
    int cp = tree.Get(p);
    while (cp > bound) {
      int i = hp.top().second;
      hp.pop();
      if (not deleted.count(i) and actives.count(i)) {
        actives.erase(i);
        tree.Sum(segs[i].first, segs[i].second, -1);
        deleted.insert(i);
        --cp;
      }
    }
  }
  cout << deleted.size() << '\n';
  for (int i : deleted) {
    cout << i + 1 << ' ';
  }
  cout << '\n';
  return 0;
}