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;
vector<int> seg[222];
int cnt[222];
int main() {
  int n, k;
  cin >> n >> k;
  int l, r;
  int max_p = 0;
  for (int i = 1; i <= n; i++) {
    cin >> l >> r;
    max_p = max(r, max_p);
    for (int j = l; j <= r; j++) {
      seg[j].push_back(i);
    }
  }
  vector<int> res;
  for (int i = 1; i <= n; i++) {
    memset(cnt, 0, sizeof(cnt));
    bool ok = true;
    for (int j = 1; j <= max_p; j++) {
      if (seg[j].size() > k) {
        ok = false;
        for (int jj = 0; jj < seg[j].size(); jj++) {
          cnt[seg[j][jj]] += seg[j].size() - k;
        }
      }
    }
    if (ok) break;
    int id = -1, tmp = 0;
    for (int j = 1; j <= n; j++) {
      if (cnt[j] > tmp) {
        id = j;
        tmp = cnt[j];
      }
    }
    res.push_back(id);
    for (int j = 1; j <= max_p; j++) {
      for (int jj = 0; jj < seg[j].size(); jj++) {
        if (seg[j][jj] == id) {
          seg[j].erase(seg[j].begin() + jj);
          break;
        }
      }
    }
  }
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++) cout << res[i] << " ";
  cout << endl;
  return 0;
}
