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;
const long long maxn = 2e5;
vector<long long> sum(4 * maxn + 5, 0);
long long get(long long v, long long l, long long r, long long id) {
  if (l > id || r <= id) return 0;
  if (l == r - 1) return sum[v];
  long long m = (l + r) / 2;
  return get(2 * v, l, m, id) + get(2 * v + 1, m, r, id) + sum[v];
}
void upd(long long v, long long l, long long r, long long tl, long long tr,
         long long id) {
  if (tl >= r || tr <= l) return;
  if (tl >= l && tr <= r) {
    sum[v] += id;
    return;
  }
  long long tm = (tl + tr) / 2;
  upd(2 * v, l, r, tl, tm, id);
  upd(2 * v + 1, l, r, tm, tr, id);
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<pair<pair<long long, long long>, long long> > a(n);
  for (long long i = 0; i < n; ++i)
    cin >> a[i].first.first >> a[i].first.second;
  for (long long i = 0; i < n; ++i) a[i].second = i;
  for (long long i = 0; i < n; ++i)
    upd(1, a[i].first.first - 1, a[i].first.second, 0, maxn, 1);
  sort(a.begin(), a.end());
  set<pair<long long, long long> > q;
  long long l = 0;
  vector<long long> ans;
  for (long long i = 0; i < maxn; ++i) {
    long long val = get(1, 0, maxn, i);
    if (val > k) {
      while (l < n && a[l].first.first - 1 <= i) {
        q.insert({maxn - a[l].first.second, a[l].second});
        ++l;
      }
      while (val > k) {
        upd(1, i, maxn - q.begin()->first, 0, maxn, -1);
        ans.push_back(q.begin()->second);
        q.erase(q.begin());
        --val;
      }
    }
  }
  cout << ans.size() << '\n';
  sort(ans.begin(), ans.end());
  for (long long i = 0; i < ans.size(); ++i) {
    cout << ans[i] + 1 << ' ';
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}