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;
bool isPowerOfTwo(int x) { return x && (!(x & (x - 1))); }
int takemod(int x) { return (x % 1000000007 + 1000000007) % 1000000007; }
int fpow(int base, int power, int MOD) {
  int result = 1;
  base = base % MOD;
  while (power) {
    if (power & 1) result = (result * base) % MOD;
    base = (base * base) % MOD;
    power = power >> 1;
  }
  return result;
}
int modinv(int x) { return fpow(x, 1000000007 - 2, 1000000007); }
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<pair<long long, long long>> A(n);
  vector<long long int> v(200005, 0);
  vector<vector<pair<long long, long long>>> ml(200005);
  for (long long int i = 0; i < (n); ++i) {
    long long int l, r;
    cin >> l >> r;
    v[l]++;
    v[r + 1]--;
    ml[l].push_back({r + 1, i + 1});
  }
  long long int cnt = 0;
  multiset<pair<long long, long long>> pr;
  vector<long long int> ans;
  for (long long int i = 0; i < (v.size()); ++i) {
    cnt += v[i];
    for (auto j : ml[i]) pr.insert(j);
    while (cnt > k) {
      cnt--;
      auto x = *pr.rbegin();
      ans.push_back(x.second);
      v[x.first]++;
      pr.erase(pr.find(x));
    }
  }
  cout << ans.size() << "\n";
  for (auto i : ans) cout << i << " ";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}