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;
const int INF = 1e9 + 100;
const long long LINF = 1e9 + 100;
const int ITER = 300;
const double EPS = 1e-6;
const long long MOD = 1e9 + 7;
const long long MAXN = 200 + 100;
int n, k;
pair<int, int> lr[MAXN];
set<int> g[MAXN];
bool intersect(pair<int, int> &a, pair<int, int> &b) {
  if (a.second < b.first || b.second < a.first) return false;
  return true;
}
bool cmp(int a, int b) { return g[a].size() < g[b].size(); }
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> lr[i].first >> lr[i].second;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (intersect(lr[i], lr[j])) {
        g[i].insert(j);
        g[j].insert(i);
      }
    }
  }
  vector<int> ans;
  deque<int> cur;
  for (int i = 0; i < n; ++i) cur.push_back(i);
  sort(cur.begin(), cur.end(), cmp);
  while (!cur.empty() && g[cur.back()].size() >= k) {
    int v = cur.back();
    cur.pop_back();
    ans.push_back(v);
    for (int i = 0; i < cur.size(); ++i) g[cur[i]].erase(v);
    sort(cur.begin(), cur.end(), cmp);
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < (int)ans.size(); ++i) cout << ans[i] + 1 << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int cs = 0; cs < t; ++cs) {
    solve();
  }
  return 0;
}
