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;
void openFiles() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  return a.second < b.second;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  vector<pair<long long int, long long int> > v;
  long long int pre[300] = {0};
  for (long long int i = 0; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    v.push_back({x, y});
    pre[x]++;
    pre[y + 1]--;
  }
  for (long long int i = 2; i <= 250; i++) {
    pre[i] += pre[i - 1];
  }
  bool done[n];
  memset(done, false, sizeof(done));
  for (long long int i = 1; i <= 250; i++) {
    while (pre[i] > k) {
      long long int pos = -1;
      for (long long int p = 0; p < n; p++) {
        if (done[p] == false && v[p].first <= i && v[p].second >= i &&
            (pos == -1 || v[p].second > v[pos].second)) {
          pos = p;
        }
      }
      if (pos != -1) {
        for (long long int j = v[pos].first; j <= v[pos].second; j++) {
          pre[j]--;
        }
      }
      done[pos] = true;
    }
  }
  long long int c = 0;
  vector<long long int> ans;
  for (long long int i = 0; i < n; i++) {
    if (done[i]) c++, ans.push_back(i);
  }
  cout << c << endl;
  for (auto x : ans) {
    cout << x + 1 << " ";
  }
  return 0;
}