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;
bool debug = false;
pair<int, int> calculate(int n, vector<tuple<int, int, int> > &V) {
  int index = -10;
  int best = 0;
  int sz = 0;
  for (int i = 0; i < n; i++) {
    int ile = 0;
    for (int j = 0; j < n; j++) {
      pair<int, int> it1 = {get<0>(V[i]), get<1>(V[i])};
      pair<int, int> it2 = {get<0>(V[j]), get<1>(V[j])};
      if (j == i) continue;
      int x1 = it1.first;
      int y1 = it1.second;
      int x2 = it2.first;
      int y2 = it2.second;
      bool b1 = x2 >= x1 and x2 <= y1;
      bool b2 = y2 >= x1 and y2 <= y1;
      bool b3 = x1 >= x2 and x1 <= y2;
      bool b4 = y1 >= x2 and y1 <= y2;
      if (b1 or b2 or b3 or b4) ile++;
    }
    if (ile > best) {
      best = ile;
      index = get<2>(V[i]);
      sz = get<1>(V[i]) - get<0>(V[i]);
    }
    if (ile == best and get<1>(V[i]) - get<0>(V[i]) > sz) {
      best = ile;
      index = get<2>(V[i]);
      sz = get<1>(V[i]) - get<0>(V[i]);
    }
  }
  if (best) best++;
  return {index, best};
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<tuple<int, int, int> > V;
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    V.push_back({l, r, i});
  }
  pair<int, int> x = calculate(n, V);
  int index = x.first;
  int best = x.second;
  vector<int> ans;
  while (best > k) {
    vector<tuple<int, int, int> > tmp;
    ans.push_back(index);
    for (auto a : V) {
      if (get<2>(a) != index) {
        tmp.push_back(a);
      }
    }
    V = tmp;
    x = calculate((int)V.size(), V);
    index = x.first;
    best = x.second;
  }
  cout << ans.size() << endl;
  for (auto a : ans) {
    cout << a + 1 << " ";
  }
  cout << endl;
  return 0;
}
