Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct var {
  int amount;
  vector<pair<int, int> > in, out;
} all[200005];
void test_case() {
  int a, b;
  cin >> a >> b;
  for (int i = 0; i < a; i++) {
    int x, y;
    cin >> x >> y;
    all[x].amount++;
    all[x].in.push_back({i, y});
    all[y + 1].amount--;
    all[y + 1].out.push_back({i, y});
  }
  int cur = 0;
  set<pair<int, int> > hold;
  vector<int> ans;
  for (int i = 0; i <= 200000; i++) {
    cur += all[i].amount;
    for (auto x : all[i].in) {
      hold.insert({-x.second, x.first});
    }
    for (auto x : all[i].out) {
      hold.erase({-x.second, x.first});
    }
    while (cur > b) {
      cur--;
      auto k = *hold.begin();
      hold.erase(hold.begin());
      all[(-k.first) + 1].amount++;
      ans.push_back(k.second);
    }
  }
  cout << ans.size() << '\n';
  for (int x : ans) cout << x + 1 << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    test_case();
  }
}