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 std::max;
using std::min;
using std::pair;
using std::string;
using std::vector;
int mass[300];
void fill(int l, int r) {
  for (l = l - 1; l < r; l++) {
    ++mass[l];
  }
}
void sub(int l, int r) {
  for (l; l <= r; l++) {
    --mass[l];
  }
}
int main() {
  int n, k;
  std::cin >> n >> k;
  vector<int> ans;
  vector<pair<int, int>> p(n);
  int m = 3000;
  for (int i = 0; i < n; i++) {
    int l, r;
    std::cin >> l >> r;
    p[i].first = l - 1;
    p[i].second = r - 1;
    fill(l, r);
    m = min(m, min(l - 1, r - 1));
  }
  for (m; m < 300; m++) {
    if (mass[m] > k) {
      while (mass[m] > k) {
        int pos_l = 20000, pos_r = -1, id = -1;
        for (int i = 0; i < n; i++) {
          if (p[i].first >= 0 && p[i].first <= m && p[i].first <= pos_l) {
            if (p[i].second > pos_r && p[i].second >= 0) {
              pos_r = p[i].second;
              pos_l = p[i].first;
              id = i;
            }
          }
        }
        if (pos_r != -1 && pos_l != 20000 && id != -1) {
          for (pos_l; pos_l <= pos_r; pos_l++) {
            --mass[pos_l];
          }
          ans.push_back(id + 1);
          p[id] = std::make_pair(-1, -1);
        } else {
          continue;
        }
      }
    }
  }
  std::cout << ans.size() << '\n';
  std::sort(ans.begin(), ans.end());
  for (int x : ans) {
    std::cout << x << " ";
  }
  return 0;
}
