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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long qpow(long long a, long long b, long long m) {
  long long r = 1;
  a %= m;
  for (; b; b >>= 1) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
  }
  return r;
}
const int inf = 0x7fffffff;
const int maxn = 2e5 + 10;
vector<pair<int, int> > g[maxn];
vector<int> ans;
set<pair<int, int> > s;
int n, k;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int l, r;
    cin >> l >> r;
    g[l].push_back(pair<int, int>(r, i));
  }
  for (int i = 1; i <= 200000; i++) {
    for (int j = 0; j < (int)(g[i]).size(); j++) s.insert(g[i][j]);
    while ((int)(s).size() > k) {
      set<pair<int, int> >::iterator it = s.end();
      it--;
      ans.push_back(it->second);
      s.erase(it);
    }
    while (!s.empty() && (s.begin()->first) <= i) s.erase(s.begin());
  }
  cout << (int)(ans).size() << endl;
  for (int i = 0; i < (int)(ans).size(); i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}