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;
set<pair<int, int>> q;
set<pair<int, int>> q1;
vector<pair<int, int>> mas[200010];
void MakeGood(int lst) {
  auto x = q.end();
  int r = q.size();
  vector<pair<int, int>> tod;
  while (r > 0) {
    r--;
    x--;
    if (x->first <= lst) {
      break;
    } else {
      tod.push_back(*x);
    }
  }
  for (auto p : tod) {
    q.erase(p);
  }
}
vector<int> ans;
int main() {
  int n, k;
  cin >> n >> k;
  int u, v;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &u, &v);
    mas[v].push_back({u, i});
  }
  for (int i = 200005; i >= 0; i--) {
    MakeGood(i);
    for (auto x : mas[i]) {
      q.insert(x);
    }
    while (q.size() > k) {
      ans.push_back((q.begin())->second + 1);
      q.erase(*(q.begin()));
    }
  }
  cout << ans.size() << endl;
  for (auto x : ans) {
    printf("%d ", x);
  }
  cout << endl;
}