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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
const char E = '\n';
const int N = 205;
const ll mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k, m[N] = {0};
  pair<int, int> a[N];
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
    for (int j = a[i].first; j <= a[i].second; j++) {
      m[j]++;
    }
  }
  vector<int> ans;
  bool used[N] = {0};
  for (int i = 1; i <= N - 5; i++) {
    while (m[i] > k) {
      int mx = -1, in = 0;
      for (int j = 1; j <= n; j++) {
        if (used[j]) continue;
        if (a[j].first <= i && a[j].second >= i && a[j].second - i > mx) {
          mx = a[j].second - i;
          in = j;
        }
      }
      ans.push_back(in);
      used[in] = 1;
      for (int j = a[in].first; j <= a[in].second; j++) {
        m[j]--;
      }
    }
  }
  cout << ans.size() << E;
  for (auto i : ans) {
    cout << i << " ";
  }
  return 0;
}