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 = 3e5 + 10;
priority_queue<pair<int, int> > que;
vector<pair<int, int> > vc[N];
int vis[N], l[N], r[N];
vector<int> ans;
int main() {
  int n, k, mx = -1;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
    mx = max(mx, r[i]);
    vc[l[i]].push_back({1, i});
    vc[r[i] + 1].push_back({-1, i});
  }
  int nw = 0;
  for (int i = 1; i <= mx; i++) {
    for (int j = 0; j < vc[i].size(); j++) {
      if (vis[vc[i][j].second]) continue;
      nw += vc[i][j].first;
      if (vc[i][j].first == -1)
        vis[vc[i][j].second] = 1;
      else
        que.push({r[vc[i][j].second], vc[i][j].second});
    }
    while (nw > k) {
      pair<int, int> pii = que.top();
      que.pop();
      if (pii.first >= i && vis[pii.second] == 0) {
        ans.push_back(pii.second);
        vis[pii.second] = 1;
        nw--;
      }
    }
  }
  printf("%d\n", ans.size());
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    printf("%d ", ans[i]);
  }
  return 0;
}