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;
const int MAXX = 1e3 + 10;
int n, k, res;
pair<pair<int, int>, pair<int, int> > a[MAXX];
vector<int> v;
bool intersect(int i, int j) {
  int l1 = a[i].first.first, l2 = a[j].first.first;
  int r1 = a[i].first.second, r2 = a[j].first.second;
  return (!(l1 > r2 || l2 > r1));
}
bool cmp(pair<pair<int, int>, pair<int, int> > x,
         pair<pair<int, int>, pair<int, int> > y) {
  return (x.second.second > y.second.second);
}
bool Ok() {
  for (int i = 0; i < n; i++)
    if (a[i].second.second > k) return 0;
  return 1;
}
void Delete() {
  v.push_back(a[0].second.first);
  res -= a[0].second.second;
  for (int i = 1; i < n; i++)
    if (intersect(i, 0)) a[i].second.second--;
  a[0].second.second = 0;
  sort(a, a + n, cmp);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++)
    cin >> a[i].first.first >> a[i].first.second, a[i].second.first = i + 1,
                                                  a[i].second.second = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && intersect(i, j)) a[i].second.second++, res++;
  sort(a, a + n, cmp);
  while (!Ok()) Delete();
  cout << v.size() << endl;
  for (auto x : v) cout << x << " ";
  return cout << endl, 0;
}
