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(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  cin >> n >> k;
  int hash[201] = {0};
  vector<pair<int, int>> v;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    v.push_back(make_pair(l, r));
    for (int j = l; j < r + 1; j++) hash[j]++;
  }
  vector<int> ans(n);
  for (int j = 0; j < 250; j++) {
    while (hash[j] > k) {
      int m = -1;
      for (int i = 0; i < n; i++) {
        if (!ans[i] && (j >= v[i].first && j <= v[i].second)) {
          if (m == -1 || v[i].second > v[m].second) m = i;
        }
      }
      if (m == -1) break;
      for (int i = v[m].first; i < v[m].second + 1; i++) hash[i]--;
      ans[m] = 1;
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << endl;
  for (int i = 0; i < n; i++)
    if (ans[i]) cout << i + 1 << " ";
}