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;
int main() {
  ios_base::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  vector<int> ox(250);
  vector<pair<int, int>> pr(n);
  for (int i = 0; i < n; ++i) {
    cin >> pr[i].first >> pr[i].second;
    ox[pr[i].first]++;
    ox[pr[i].second + 1]--;
  }
  for (int i = 1; i < 250; ++i) {
    ox[i] += ox[i - 1];
  }
  vector<int> ans(n);
  for (int i = 0; i < 250; ++i) {
    while (ox[i] > k) {
      int pos = -1;
      for (int p = 0; p < n; ++p) {
        if (!ans[p] && (pr[p].first <= i && i <= pr[p].second) &&
            (pos == -1 || pr[p].second > pr[pos].second)) {
          pos = p;
        }
      }
      assert(pos != -1);
      for (int j = pr[pos].first; j <= pr[pos].second; ++j) {
        --ox[j];
      }
      ans[pos] = 1;
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << "\n";
  for (int i = 0; i < n; ++i) {
    if (ans[i]) cout << (i + 1) << " ";
  }
  cout << "\n";
  return 0;
}