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;
const int N = 225;
const int M = 1e9 + 7;
int st[N];
vector<pair<int, int> > segs;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    segs.push_back({x, y});
    st[x]++;
    st[y + 1]--;
  }
  for (int i = 0; i + 1 < N; i++) {
    st[i + 1] += st[i];
  }
  vector<int> ans(n, 0);
  for (int i = 1; i < N; i++) {
    while (st[i] > k) {
      int pos = -1;
      for (int j = 0; j < n; j++) {
        if (!ans[j] && (segs[j].first <= i && i <= segs[j].second) &&
            (pos == -1 || segs[pos].second < segs[j].second)) {
          pos = j;
        }
      }
      if (pos > -1) {
        for (int j = segs[pos].first; j <= segs[pos].second; j++) st[j]--;
        ans[pos] = 1;
      }
    }
  }
  cout << accumulate(ans.begin(), ans.end(), 0) << '\n';
  for (int i = 0; i < n; i++) {
    if (ans[i]) cout << i + 1 << ' ';
  }
  return 0;
}