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 
#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5;
const int INF = 1e9 + 1e5 + 2;
const int mod = 1e9 + 7;
long long pref[40];
long long po[40];
int main() {
  int n, k;
  cin >> n >> k;
  vector<pair<int, int> > v;
  int x, y, mi = 1000, ma = -1;
  for (int i = 0; i < n; ++i) {
    cin >> x >> y;
    mi = min(x, mi);
    ma = max(y, ma);
    v.push_back({x, y});
  }
  vector<int> ans;
  map<int, int> mp;
  for (int i = mi; i <= ma; ++i) {
    vector<pair<int, int> > p;
    for (int j = 0; j < v.size(); ++j) {
      if (i >= v[j].first && i <= v[j].second)
        p.push_back({v[j].second - v[j].first, j + 1});
    }
    if (p.size() > k) {
      sort(p.rbegin(), p.rend());
      for (int j = 0; j < p.size() - k; ++j) {
        if (mp[p[j].second] == 0) {
          ans.push_back(p[j].second);
          mp[p[j].second] = 1;
        }
      }
    }
  }
  cout << ans.size() << "\n";
  for (auto a : ans) cout << a << " ";
}
