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;
bool boom[300];
int main() {
  ios::sync_with_stdio(false);
  long long t, n, k, i, x, y;
  vector<long long> ans;
  cin >> n >> k;
  long long cum[300] = {0};
  vector<pair<int, int> > v;
  for (i = 0; i < n; i++) {
    cin >> x >> y;
    v.push_back({x, y});
    cum[x]++;
    cum[y + 1]--;
  }
  for (i = 1; i < 300; i++) {
    cum[i] = cum[i - 1] + cum[i];
    if (cum[i] > k) {
      boom[i] = true;
    }
  }
  for (int kk = 0; kk < 300; kk++) {
    long long mx = 0, idx = 0;
    for (i = 1; i < 300; i++) {
      boom[i] = cum[i] > k;
    }
    for (i = 0; i < v.size(); i++) {
      int cnt = 0;
      for (int j = v[i].first; j <= v[i].second; j++) {
        if (boom[j]) {
          cnt++;
        }
      }
      if (cnt > mx) {
        mx = cnt;
        idx = i;
      }
    }
    if (mx == 0) break;
    ans.push_back(idx);
    for (i = v[idx].first; i <= v[idx].second; i++) {
      cum[i]--;
    }
    v[idx].first = 0;
    v[idx].second = 0;
  }
  cout << ans.size() << "\n";
  for (i = 0; i < ans.size(); i++) {
    cout << ans[i] + 1 << " ";
  }
  return 0;
}
