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 swap(long long &a, long long &b) {
  auto tm = a;
  a = b;
  b = tm;
}
const long long N = 2e5 + 5;
const long long INF = 1e16;
long long bit[N] = {0};
void add(long long idx, long long val) {
  for (++idx; idx < N; idx += idx & -idx) bit[idx] += val;
}
void range_add(long long l, long long r, long long val) {
  add(l, val);
  add(r + 1, -val);
}
long long point_query(long long idx) {
  long long ret = 0;
  for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
  return ret;
}
vector<long long> st[N], ed[N];
void bheja_fry() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > v(n);
  for (long long(i) = (0); (i) < (n); (i)++) {
    cin >> v[i].first >> v[i].second;
    st[v[i].first].push_back(i);
    ed[v[i].second].push_back(i);
    range_add(v[i].first, v[i].second, 1);
  }
  long long ct = 0;
  set<pair<long long, long long> > s;
  vector<long long> ans;
  for (long long l = 1; l <= 2e5; l++) {
    ct = point_query(l);
    for (auto &(i) : (st[l])) s.insert({v[i].second, i});
    while (ct > k) {
      long long id = (*(--s.end())).second;
      ans.push_back(id);
      ct--;
      range_add(v[id].first, v[id].second, -1);
      s.erase((--s.end()));
    }
    for (auto &(i) : (ed[l])) s.erase({v[i].second, i});
  }
  assert(!(long long)(s).size());
  cout << (long long)(ans).size() << "\n";
  for (auto &(i) : (ans)) cout << i + 1 << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long(i) = (1); (i) < (t + 1); (i)++) {
    bheja_fry();
    if (i < t) cout << "\n";
  }
  cerr << (1000 * clock()) / CLOCKS_PER_SEC << "ms";
  return 0;
}