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 double eps = 1e-16;
const long long MOD = (int)998244353;
const int INF = 2 * (int)1e9 + 3;
const int NHASH = 2;
const int MAXN = 200001;
inline long long mod(long long x, long long m = MOD) {
  if (x < 0) x += m;
  return (x % m);
}
int n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  vector<pair<int, int> > pt;
  vector<pair<int, int> > s(n + 1);
  for (int i = 1; i <= n; ++i) {
    int l, r;
    cin >> l >> r;
    s[i] = {l, r};
    pt.push_back({l, -i});
    pt.push_back({r, i});
  }
  sort(pt.begin(), pt.end());
  ;
  vector<int> ans;
  vector<int> exc(n + 1, 0);
  priority_queue<pair<int, int> > q;
  int cur = 0;
  for (int i = 0; i < 2 * n; ++i) {
    int ind = -pt[i].second;
    if (ind > 0) {
      cur++;
      q.push({s[ind].second, ind});
      if (cur > k) {
        int ind2 = q.top().second;
        ans.push_back(ind2);
        q.pop();
        cur--;
        exc[ind2] = 1;
      }
    } else if (!exc[-ind]) {
      cur--;
    }
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << " ";
  }
  return 0;
}