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;
int fx[] = {0, 0, 1, -1, -1, 1, -1, 1};
int fy[] = {1, -1, 0, 0, -1, 1, 1, -1};
vector<pair<int, int> > vec;
int arr[204];
bool mark[204];
int main() {
  int n, a, b, k;
  scanf("%d%*c", &n), scanf("%d%*c", &k);
  for (int i = 0; i < n; i++) {
    scanf("%d%*c", &a), scanf("%d%*c", &b);
    vec.push_back({a, b});
    for (int j = a; j <= b; j++) arr[j]++;
  }
  vector<int> ans;
  while (1) {
    int mx = 0, in;
    for (int j = 0; j < vec.size(); j++) {
      int co = 0;
      if (mark[j] == 1) continue;
      for (int i = 1; i < 204; i++) {
        if (arr[i] > k && i >= vec[j].first && i <= vec[j].second) {
          co++;
        }
      }
      if (co > mx) {
        mx = co;
        in = j;
      }
    }
    if (mx == 0)
      break;
    else {
      ans.push_back(in + 1);
      mark[in] = 1;
      for (int i = vec[in].first; i <= vec[in].second; i++) arr[i]--;
    }
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
}
